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#include <bits/stdc++.h>
int main() {
int a, b, c, m;
scanf("%d %d %d %d", &a, &b, &c, &m);
if (2 * c < m || c > m * 2) {
c = -1;
} else if (c < m && 2 * c >= m)
c = m;
else if (c >= m && c <= m * 2)
c = c;
if (2 * b <= 2 * m) {
b = -1;
} else
b = 2 * b;
if (2 * a <= 2 * m) {
a = -1;
} else
a = 2 * a;
if (a == -1 || b == -1 || c == -1)
printf("-1\n");
else
printf("%d\n%d\n%d\n", a, b, c);
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int a, b, c, m;
scanf("%d %d %d %d", &a, &b, &c, &m);
if (2 * c < m || c > m * 2) {
c = -1;
} else if (c < m && 2 * c >= m)
c = m;
else if (c >= m && c <= m * 2)
c = c;
if (2 * b <= 2 * m) {
b = -1;
} else
b = 2 * b;
if (2 * a <= 2 * m) {
a = -1;
} else
a = 2 * a;
if (a == -1 || b == -1 || c == -1)
printf("-1\n");
else
printf("%d\n%d\n%d\n", a, b, c);
}
``` |
#include <bits/stdc++.h>
using namespace std;
void scanint(int &x) {
int c = getchar_unlocked();
x = 0;
while (((c < 48 || c > 57))) c = getchar_unlocked();
for (; c > 47 && c < 58; c = getchar_unlocked()) {
x = (x << 1) + (x << 3) + c - 48;
}
}
int v1, v2, v3, vm, x1, x2, x3;
int main() {
cin >> v1 >> v2 >> v3 >> vm;
if (2 * vm < v3 || vm > 2 * v3) {
cout << "-1\n";
getchar();
getchar();
return 0;
}
if (2 * max(v3, vm) >= 2 * v2 || 2 * max(v3, vm) >= 2 * v1) {
cout << "-1\n";
getchar();
getchar();
return 0;
}
cout << 2 * v1 << endl << 2 * v2 << endl << max(v3, vm) << endl;
getchar();
getchar();
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void scanint(int &x) {
int c = getchar_unlocked();
x = 0;
while (((c < 48 || c > 57))) c = getchar_unlocked();
for (; c > 47 && c < 58; c = getchar_unlocked()) {
x = (x << 1) + (x << 3) + c - 48;
}
}
int v1, v2, v3, vm, x1, x2, x3;
int main() {
cin >> v1 >> v2 >> v3 >> vm;
if (2 * vm < v3 || vm > 2 * v3) {
cout << "-1\n";
getchar();
getchar();
return 0;
}
if (2 * max(v3, vm) >= 2 * v2 || 2 * max(v3, vm) >= 2 * v1) {
cout << "-1\n";
getchar();
getchar();
return 0;
}
cout << 2 * v1 << endl << 2 * v2 << endl << max(v3, vm) << endl;
getchar();
getchar();
return 0;
}
``` |
#include <bits/stdc++.h>
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 1e5 + 5;
const double eps = 1e-9;
using namespace std;
int arr[MAXN];
int main() {
for (int i = 0; i < 4; i++) scanf("%d", arr + i);
for (int a = 0; a <= 300; a++) {
if (arr[2] > a || arr[3] > a) continue;
if (2 * arr[2] < a || 2 * arr[3] < a) continue;
for (int b = a + 1; b <= 300; b++) {
if (arr[1] > b) continue;
if (2 * arr[1] < b || 2 * arr[3] >= b) continue;
for (int c = b + 1; c <= 300; c++) {
if (arr[0] > c) continue;
if (2 * arr[0] < c || 2 * arr[3] >= c) continue;
printf("%d\n%d\n%d\n", c, b, a);
return 0;
}
}
}
printf("-1\n");
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 1e5 + 5;
const double eps = 1e-9;
using namespace std;
int arr[MAXN];
int main() {
for (int i = 0; i < 4; i++) scanf("%d", arr + i);
for (int a = 0; a <= 300; a++) {
if (arr[2] > a || arr[3] > a) continue;
if (2 * arr[2] < a || 2 * arr[3] < a) continue;
for (int b = a + 1; b <= 300; b++) {
if (arr[1] > b) continue;
if (2 * arr[1] < b || 2 * arr[3] >= b) continue;
for (int c = b + 1; c <= 300; c++) {
if (arr[0] > c) continue;
if (2 * arr[0] < c || 2 * arr[3] >= c) continue;
printf("%d\n%d\n%d\n", c, b, a);
return 0;
}
}
}
printf("-1\n");
return 0;
}
``` |
#include <bits/stdc++.h>
int V[10], v, flag;
int main() {
scanf("%d%d%d%d", &V[1], &V[2], &V[3], &v);
for (int c1 = V[1]; c1 <= 2 * V[1]; c1++)
for (int c2 = V[2]; c2 <= 2 * V[2]; c2++)
for (int c3 = V[3]; c3 <= 2 * V[3]; c3++) {
if (c1 > c2 && c2 > c3 && c1 > 2 * v && c2 > 2 * v && c3 >= v &&
2 * v >= c3) {
printf("%d\n%d\n%d", c1, c2, c3);
return 0;
}
}
printf("-1");
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
int V[10], v, flag;
int main() {
scanf("%d%d%d%d", &V[1], &V[2], &V[3], &v);
for (int c1 = V[1]; c1 <= 2 * V[1]; c1++)
for (int c2 = V[2]; c2 <= 2 * V[2]; c2++)
for (int c3 = V[3]; c3 <= 2 * V[3]; c3++) {
if (c1 > c2 && c2 > c3 && c1 > 2 * v && c2 > 2 * v && c3 >= v &&
2 * v >= c3) {
printf("%d\n%d\n%d", c1, c2, c3);
return 0;
}
}
printf("-1");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int anss[3];
int main() {
int V1, V2, V3, Vm, minn, maxx;
bool ans;
ans = true;
scanf("%d%d%d%d", &V1, &V2, &V3, &Vm);
minn = 0;
maxx = 99999;
minn = max(minn, Vm);
maxx = min(maxx, 2 * Vm);
minn = max(minn, V3);
maxx = min(maxx, 2 * V3);
if (minn > maxx)
ans = false;
else {
anss[2] = minn;
minn = anss[2] + 1;
maxx = 99999;
minn = max(minn, Vm);
minn = max(minn, 2 * Vm + 1);
minn = max(minn, V2);
maxx = min(maxx, 2 * V2);
if (minn > maxx)
ans = false;
else {
anss[1] = minn;
minn = anss[1] + 1;
maxx = 99999;
minn = max(minn, Vm);
minn = max(minn, 2 * Vm + 1);
minn = max(minn, V1);
maxx = min(maxx, 2 * V1);
if (minn > maxx)
ans = false;
else
anss[0] = minn;
}
}
if (ans)
for (int i = 0; i < 3; i++) printf("%d\n", anss[i]);
else
printf("-1\n");
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int anss[3];
int main() {
int V1, V2, V3, Vm, minn, maxx;
bool ans;
ans = true;
scanf("%d%d%d%d", &V1, &V2, &V3, &Vm);
minn = 0;
maxx = 99999;
minn = max(minn, Vm);
maxx = min(maxx, 2 * Vm);
minn = max(minn, V3);
maxx = min(maxx, 2 * V3);
if (minn > maxx)
ans = false;
else {
anss[2] = minn;
minn = anss[2] + 1;
maxx = 99999;
minn = max(minn, Vm);
minn = max(minn, 2 * Vm + 1);
minn = max(minn, V2);
maxx = min(maxx, 2 * V2);
if (minn > maxx)
ans = false;
else {
anss[1] = minn;
minn = anss[1] + 1;
maxx = 99999;
minn = max(minn, Vm);
minn = max(minn, 2 * Vm + 1);
minn = max(minn, V1);
maxx = min(maxx, 2 * V1);
if (minn > maxx)
ans = false;
else
anss[0] = minn;
}
}
if (ans)
for (int i = 0; i < 3; i++) printf("%d\n", anss[i]);
else
printf("-1\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d;
int main() {
cin.sync_with_stdio(false), cin.tie(0);
cin >> a >> b >> c >> d;
for (long long i = 1; i <= 200; ++i) {
for (long long j = 1; j <= 200; ++j) {
for (long long k = 1; k <= 200; ++k) {
if (i <= j || j <= k) continue;
if (i >= a && i <= 2 * a && j >= b && j <= 2 * b && k >= c &&
k <= 2 * c) {
if (k >= d && 2 * d >= k && 2 * d < i && 2 * d < j) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return 0;
}
}
}
}
}
cout << -1 << endl;
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d;
int main() {
cin.sync_with_stdio(false), cin.tie(0);
cin >> a >> b >> c >> d;
for (long long i = 1; i <= 200; ++i) {
for (long long j = 1; j <= 200; ++j) {
for (long long k = 1; k <= 200; ++k) {
if (i <= j || j <= k) continue;
if (i >= a && i <= 2 * a && j >= b && j <= 2 * b && k >= c &&
k <= 2 * c) {
if (k >= d && 2 * d >= k && 2 * d < i && 2 * d < j) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return 0;
}
}
}
}
}
cout << -1 << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
int a1, b1, c1;
scanf("%d %d %d %d", &a, &b, &c, &d);
if (d > 2 * c || c > 2 * d) {
printf("-1\n");
return 0;
}
c1 = max(c, d);
b1 = max(c1 + 1, b);
if (b1 > 2 * b) {
printf("-1\n");
return 0;
}
if (2 * d > 2 * b) {
printf("-1\n");
return 0;
}
b1 = max(b1, 2 * d + 1);
if (b1 > 2 * b) {
printf("-1\n");
return 0;
}
a1 = max(a, b1 + 1);
if (a1 > 2 * a) {
printf("-1\n");
return 0;
}
if (2 * d > 2 * a) {
printf("-1\n");
return 0;
}
a1 = max(a1, 2 * d + 1);
if (a1 > 2 * a) {
printf("-1\n");
return 0;
}
printf("%d\n%d\n%d\n", a1, b1, c1);
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
int a1, b1, c1;
scanf("%d %d %d %d", &a, &b, &c, &d);
if (d > 2 * c || c > 2 * d) {
printf("-1\n");
return 0;
}
c1 = max(c, d);
b1 = max(c1 + 1, b);
if (b1 > 2 * b) {
printf("-1\n");
return 0;
}
if (2 * d > 2 * b) {
printf("-1\n");
return 0;
}
b1 = max(b1, 2 * d + 1);
if (b1 > 2 * b) {
printf("-1\n");
return 0;
}
a1 = max(a, b1 + 1);
if (a1 > 2 * a) {
printf("-1\n");
return 0;
}
if (2 * d > 2 * a) {
printf("-1\n");
return 0;
}
a1 = max(a1, 2 * d + 1);
if (a1 > 2 * a) {
printf("-1\n");
return 0;
}
printf("%d\n%d\n%d\n", a1, b1, c1);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e9;
const long double PI = 3.14159265358979323846;
const long long INF = 1e6 - 1;
long long prime1(long long n) {
bool ans = false;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
ans = true;
break;
}
}
if (ans == 0) {
return 1;
} else {
return 0;
}
}
long long Catalan(long long n) {
long long i, sum;
if (n <= 0) {
return 1;
}
sum = 0;
for (i = 0; i < n; i++) {
sum += Catalan(i) * Catalan((n - 1) - i);
}
return sum;
}
using namespace std;
int main() {
int a, b, c, n;
cin >> a >> b >> c >> n;
if (2 * c >= n && n >= c && b > n && a > n)
cout << a * 2 << " " << b * 2 << " " << n;
else if (2 * n >= c && c >= n)
cout << a * 2 << " " << b * 2 << " " << c;
else
cout << -1;
}
| ### 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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e9;
const long double PI = 3.14159265358979323846;
const long long INF = 1e6 - 1;
long long prime1(long long n) {
bool ans = false;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
ans = true;
break;
}
}
if (ans == 0) {
return 1;
} else {
return 0;
}
}
long long Catalan(long long n) {
long long i, sum;
if (n <= 0) {
return 1;
}
sum = 0;
for (i = 0; i < n; i++) {
sum += Catalan(i) * Catalan((n - 1) - i);
}
return sum;
}
using namespace std;
int main() {
int a, b, c, n;
cin >> a >> b >> c >> n;
if (2 * c >= n && n >= c && b > n && a > n)
cout << a * 2 << " " << b * 2 << " " << n;
else if (2 * n >= c && c >= n)
cout << a * 2 << " " << b * 2 << " " << c;
else
cout << -1;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v0, v1, v2, vm, a0, a1, a2;
bool err = true;
scanf("%d%d%d%d", &v2, &v1, &v0, &vm);
for (int i = v0; i <= v0 * 2; i++) {
for (int j = v1; j <= v1 * 2; j++) {
for (int k = v2; k <= v2 * 2; k++) {
if (vm <= i && vm <= j && vm <= k && i < j && j < k && vm * 2 >= i &&
vm * 2 < j) {
a0 = k;
a1 = j;
a2 = i;
err = false;
break;
}
}
}
}
if (err) {
printf("-1\n");
} else {
printf("%d\n%d\n%d\n", a0, a1, a2);
}
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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v0, v1, v2, vm, a0, a1, a2;
bool err = true;
scanf("%d%d%d%d", &v2, &v1, &v0, &vm);
for (int i = v0; i <= v0 * 2; i++) {
for (int j = v1; j <= v1 * 2; j++) {
for (int k = v2; k <= v2 * 2; k++) {
if (vm <= i && vm <= j && vm <= k && i < j && j < k && vm * 2 >= i &&
vm * 2 < j) {
a0 = k;
a1 = j;
a2 = i;
err = false;
break;
}
}
}
}
if (err) {
printf("-1\n");
} else {
printf("%d\n%d\n%d\n", a0, a1, a2);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
vector<int> vec(5);
for (int i = 0; i < 4; i++) cin >> vec[i];
for (int i = vec[0]; i <= 200; i++) {
if (2 * vec[0] < i) continue;
for (int j = vec[1]; j <= 200; j++) {
if (2 * vec[1] < j) continue;
for (int k = vec[2]; k <= 200; k++) {
if (i > j && j > k) {
if (2 * vec[2] < k || 2 * vec[3] < k) continue;
if (vec[3] <= i && vec[3] <= j && vec[3] <= k) {
if (vec[3] * 2 < i && vec[3] * 2 < j)
return cout << i << endl << j << endl << k, 0;
}
}
}
}
}
cout << -1;
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
vector<int> vec(5);
for (int i = 0; i < 4; i++) cin >> vec[i];
for (int i = vec[0]; i <= 200; i++) {
if (2 * vec[0] < i) continue;
for (int j = vec[1]; j <= 200; j++) {
if (2 * vec[1] < j) continue;
for (int k = vec[2]; k <= 200; k++) {
if (i > j && j > k) {
if (2 * vec[2] < k || 2 * vec[3] < k) continue;
if (vec[3] <= i && vec[3] <= j && vec[3] <= k) {
if (vec[3] * 2 < i && vec[3] * 2 < j)
return cout << i << endl << j << endl << k, 0;
}
}
}
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v;
cin >> v1 >> v2 >> v3 >> v;
if (v > 2 * v3 || v3 > 2 * v || v >= v2)
cout << -1 << "\n";
else {
int a = max(v, v3);
cout << 2 * v1 << " " << 2 * v2 << " " << a << "\n";
}
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v;
cin >> v1 >> v2 >> v3 >> v;
if (v > 2 * v3 || v3 > 2 * v || v >= v2)
cout << -1 << "\n";
else {
int a = max(v, v3);
cout << 2 * v1 << " " << 2 * v2 << " " << a << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
;
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
int a1, a2, a3;
if (max(v3, vm) <= min(vm * 2, v3 * 2))
a3 = max(v3, vm);
else
return puts("-1");
int min2 = max(a3 + 1, max(v3, max(v2, vm * 2 + 1)));
int max2 = v2 * 2;
if (min2 <= max2)
a2 = min2;
else
return puts("-1");
int min1 = max(a2 + 1, max(v1, max(v2, vm * 2 + 1)));
int max1 = v1 * 2;
if (min1 <= max1)
a1 = min1;
else
return puts("-1");
cout << a1 << " " << a2 << " " << a3 << "\n";
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
;
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
int a1, a2, a3;
if (max(v3, vm) <= min(vm * 2, v3 * 2))
a3 = max(v3, vm);
else
return puts("-1");
int min2 = max(a3 + 1, max(v3, max(v2, vm * 2 + 1)));
int max2 = v2 * 2;
if (min2 <= max2)
a2 = min2;
else
return puts("-1");
int min1 = max(a2 + 1, max(v1, max(v2, vm * 2 + 1)));
int max1 = v1 * 2;
if (min1 <= max1)
a1 = min1;
else
return puts("-1");
cout << a1 << " " << a2 << " " << a3 << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
int flag;
while (cin >> v1 >> v2 >> v3 >> vm) {
flag = 0;
int i, j, k;
for (i = 1; i <= 200; i++) {
for (j = 1; j < i; j++) {
for (k = 1; k < j; k++) {
if (2 * v1 >= i && v1 <= i && v2 <= j && 2 * v2 >= j && v3 <= k &&
2 * v3 >= k && 2 * vm >= k && vm <= k && 2 * vm < j) {
flag = 1;
break;
}
}
if (flag) break;
}
if (flag) break;
}
if (flag) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
} else {
cout << "-1" << endl;
}
}
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
int flag;
while (cin >> v1 >> v2 >> v3 >> vm) {
flag = 0;
int i, j, k;
for (i = 1; i <= 200; i++) {
for (j = 1; j < i; j++) {
for (k = 1; k < j; k++) {
if (2 * v1 >= i && v1 <= i && v2 <= j && 2 * v2 >= j && v3 <= k &&
2 * v3 >= k && 2 * vm >= k && vm <= k && 2 * vm < j) {
flag = 1;
break;
}
}
if (flag) break;
}
if (flag) break;
}
if (flag) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
} else {
cout << "-1" << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, ans1, ans2, ans3;
int main() {
cin >> a >> b >> c >> d;
ans1 = max(a, 2 * d + 1);
ans2 = max(b, 2 * d + 1);
ans3 = max(c, d);
if (ans2 == ans3) {
if (ans1 == ans2) {
ans1 += 2;
}
ans2++;
} else if (ans1 == ans2)
ans1++;
if (ans1 > 2 * a) {
cout << -1 << endl;
} else if (ans2 > 2 * b) {
cout << -1 << endl;
} else if (ans3 > 2 * c) {
cout << -1 << endl;
} else if (ans3 > 2 * d) {
cout << -1 << endl;
} else
cout << ans1 << "\n" << ans2 << "\n" << ans3 << endl;
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, ans1, ans2, ans3;
int main() {
cin >> a >> b >> c >> d;
ans1 = max(a, 2 * d + 1);
ans2 = max(b, 2 * d + 1);
ans3 = max(c, d);
if (ans2 == ans3) {
if (ans1 == ans2) {
ans1 += 2;
}
ans2++;
} else if (ans1 == ans2)
ans1++;
if (ans1 > 2 * a) {
cout << -1 << endl;
} else if (ans2 > 2 * b) {
cout << -1 << endl;
} else if (ans3 > 2 * c) {
cout << -1 << endl;
} else if (ans3 > 2 * d) {
cout << -1 << endl;
} else
cout << ans1 << "\n" << ans2 << "\n" << ans3 << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
inline long long readll() {
int sign = 1;
char c;
for (c = getchar(); !isdigit(c); c = getchar())
if (c == '-') sign = -sign;
long long res = c - '0';
for (c = getchar(); isdigit(c); c = getchar()) res = res * 10 + c - '0';
return sign * res;
}
void writell(long long x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) writell(x / 10);
putchar(x % 10 + '0');
}
void inp() { cin >> v1 >> v2 >> v3 >> vm; }
void out() {
if (vm < v2 && vm <= 2 * v3 && v3 <= 2 * vm) {
writell(v1 * 2);
putchar('\n');
writell(v2 * 2);
putchar('\n');
writell(min(v3 * 2, vm * 2));
} else {
putchar('-');
putchar('1');
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
inp();
out();
return 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
inline long long readll() {
int sign = 1;
char c;
for (c = getchar(); !isdigit(c); c = getchar())
if (c == '-') sign = -sign;
long long res = c - '0';
for (c = getchar(); isdigit(c); c = getchar()) res = res * 10 + c - '0';
return sign * res;
}
void writell(long long x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) writell(x / 10);
putchar(x % 10 + '0');
}
void inp() { cin >> v1 >> v2 >> v3 >> vm; }
void out() {
if (vm < v2 && vm <= 2 * v3 && v3 <= 2 * vm) {
writell(v1 * 2);
putchar('\n');
writell(v2 * 2);
putchar('\n');
writell(min(v3 * 2, vm * 2));
} else {
putchar('-');
putchar('1');
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
inp();
out();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int minx = -1, miny = -1, minz = -1, maxx = 1000, maxy = 1000, maxz = 1000;
int a, b, c, d;
cin >> a >> b >> c >> d;
minz = max(minz, c);
minz = max(minz, d);
maxz = min(maxz, 2 * c);
maxz = min(maxz, 2 * d);
miny = max(miny, b);
miny = max(miny, 2 * d + 1);
miny = max(miny, minz + 1);
maxy = min(maxy, 2 * b);
minx = max(minx, a);
minx = max(minx, 2 * d + 1);
minx = max(minx, miny + 1);
maxx = min(maxx, 2 * a);
if (maxx < minx || maxy < miny || maxz < minz) {
cout << "-1\n";
return 0;
}
cout << minx << "\n" << miny << "\n" << minz << "\n";
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int minx = -1, miny = -1, minz = -1, maxx = 1000, maxy = 1000, maxz = 1000;
int a, b, c, d;
cin >> a >> b >> c >> d;
minz = max(minz, c);
minz = max(minz, d);
maxz = min(maxz, 2 * c);
maxz = min(maxz, 2 * d);
miny = max(miny, b);
miny = max(miny, 2 * d + 1);
miny = max(miny, minz + 1);
maxy = min(maxy, 2 * b);
minx = max(minx, a);
minx = max(minx, 2 * d + 1);
minx = max(minx, miny + 1);
maxx = min(maxx, 2 * a);
if (maxx < minx || maxy < miny || maxz < minz) {
cout << "-1\n";
return 0;
}
cout << minx << "\n" << miny << "\n" << minz << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if (c * 2 < d || d >= b)
cout << -1 << endl;
else if (d * 2 < c)
cout << -1 << endl;
else {
cout << a * 2 << endl;
cout << b * 2 << endl;
cout << max(c, d) << endl;
}
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if (c * 2 < d || d >= b)
cout << -1 << endl;
else if (d * 2 < c)
cout << -1 << endl;
else {
cout << a * 2 << endl;
cout << b * 2 << endl;
cout << max(c, d) << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int V1, V2, V3, Vm;
bool OK(int x) {
if (Vm <= x && 2 * Vm >= x) {
return true;
}
return false;
}
void solve() {
int i, j, k;
for (i = V1; i <= 2 * V1; ++i) {
int r;
if (2 * V2 < i)
r = 2 * V2;
else
r = i - 1;
for (j = V2; j <= r; ++j) {
int rr;
if (2 * V3 < j)
rr = 2 * V3;
else
rr = j - 1;
for (k = V3; k <= rr; ++k) {
if (!OK(i) && !OK(j) && OK(k)) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return;
}
}
}
}
cout << -1 << endl;
}
int main(void) {
cin >> V1 >> V2 >> V3 >> Vm;
solve();
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int V1, V2, V3, Vm;
bool OK(int x) {
if (Vm <= x && 2 * Vm >= x) {
return true;
}
return false;
}
void solve() {
int i, j, k;
for (i = V1; i <= 2 * V1; ++i) {
int r;
if (2 * V2 < i)
r = 2 * V2;
else
r = i - 1;
for (j = V2; j <= r; ++j) {
int rr;
if (2 * V3 < j)
rr = 2 * V3;
else
rr = j - 1;
for (k = V3; k <= rr; ++k) {
if (!OK(i) && !OK(j) && OK(k)) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return;
}
}
}
}
cout << -1 << endl;
}
int main(void) {
cin >> V1 >> V2 >> V3 >> Vm;
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void ans(int64_t n) {
cout << n;
exit(0);
}
int32_t main() {
int64_t a, b, c, d;
cin >> a >> b >> c >> d;
int64_t zl = max(c, d), zr = 2 * min(c, d);
if (zl > zr) ans(-1);
int64_t z = zl, yl = max(z + 1, max(2 * d + 1, b)), yr = 2 * b;
if (yl > yr) ans(-1);
int64_t y = yl;
int64_t xl = max(y + 1, a);
int64_t xr = 2 * a;
int64_t x = xl;
if (xl > xr) ans(-1);
cout << x << endl << y << endl << z;
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void ans(int64_t n) {
cout << n;
exit(0);
}
int32_t main() {
int64_t a, b, c, d;
cin >> a >> b >> c >> d;
int64_t zl = max(c, d), zr = 2 * min(c, d);
if (zl > zr) ans(-1);
int64_t z = zl, yl = max(z + 1, max(2 * d + 1, b)), yr = 2 * b;
if (yl > yr) ans(-1);
int64_t y = yl;
int64_t xl = max(y + 1, a);
int64_t xr = 2 * a;
int64_t x = xl;
if (xl > xr) ans(-1);
cout << x << endl << y << endl << z;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int sz1 = a; sz1 <= 2 * a; sz1++) {
for (int sz2 = b; sz2 <= 2 * b; sz2++) {
for (int sz3 = c; sz3 <= 2 * c; sz3++) {
if (!(sz1 > sz2 and sz2 > sz3)) continue;
if (d <= sz1 and d <= sz2 and d <= sz3 and (2 * d) >= sz3 and
(2 * d) < sz2 and (2 * d) < sz1) {
cout << sz1 << endl << sz2 << endl << sz3;
return 0;
}
}
}
}
cout << -1;
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int sz1 = a; sz1 <= 2 * a; sz1++) {
for (int sz2 = b; sz2 <= 2 * b; sz2++) {
for (int sz3 = c; sz3 <= 2 * c; sz3++) {
if (!(sz1 > sz2 and sz2 > sz3)) continue;
if (d <= sz1 and d <= sz2 and d <= sz3 and (2 * d) >= sz3 and
(2 * d) < sz2 and (2 * d) < sz1) {
cout << sz1 << endl << sz2 << endl << sz3;
return 0;
}
}
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int m, n, o, p, flag = 0;
cin >> m >> n >> o >> p;
for (int i = 1; i <= 200; i++) {
for (int k = 1; k < i; k++) {
for (int j = 1; j < i; j++) {
if (m <= i && 2 * m >= i && n <= k && 2 * n >= k && o <= j &&
2 * o >= j && p <= j && 2 * p >= j && 2 * p < k) {
cout << i << endl << k << endl << j << endl;
flag = 1;
return 0;
}
}
}
}
if (flag == 0) cout << "-1" << endl;
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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int m, n, o, p, flag = 0;
cin >> m >> n >> o >> p;
for (int i = 1; i <= 200; i++) {
for (int k = 1; k < i; k++) {
for (int j = 1; j < i; j++) {
if (m <= i && 2 * m >= i && n <= k && 2 * n >= k && o <= j &&
2 * o >= j && p <= j && 2 * p >= j && 2 * p < k) {
cout << i << endl << k << endl << j << endl;
flag = 1;
return 0;
}
}
}
}
if (flag == 0) cout << "-1" << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm, a, b, c;
scanf("%d%d%d%d", &v1, &v2, &v3, &vm);
a = 2 * v1, b = 2 * v2;
if (vm >= v1 || vm >= v2 || max(vm, v3) > 2 * min(vm, v3)) {
printf("-1\n");
} else
printf("%d\n%d\n%d\n", a, b, 2 * min(vm, v3));
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm, a, b, c;
scanf("%d%d%d%d", &v1, &v2, &v3, &vm);
a = 2 * v1, b = 2 * v2;
if (vm >= v1 || vm >= v2 || max(vm, v3) > 2 * min(vm, v3)) {
printf("-1\n");
} else
printf("%d\n%d\n%d\n", a, b, 2 * min(vm, v3));
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int s1, s2, s3, v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
s1 = max(v1, 2 * vm + 2);
s2 = max(v2, 2 * vm + 1);
s3 = max(v3, vm);
if (s1 <= 2 * v1 && s2 <= 2 * v2 && s3 <= 2 * v3 && s1 > 2 * vm &&
s2 > 2 * vm && s3 <= 2 * vm)
cout << s1 << endl << s2 << endl << s3 << endl;
else
cout << -1 << endl;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int s1, s2, s3, v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
s1 = max(v1, 2 * vm + 2);
s2 = max(v2, 2 * vm + 1);
s3 = max(v3, vm);
if (s1 <= 2 * v1 && s2 <= 2 * v2 && s3 <= 2 * v3 && s1 > 2 * vm &&
s2 > 2 * vm && s3 <= 2 * vm)
cout << s1 << endl << s2 << endl << s3 << endl;
else
cout << -1 << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
while (scanf("%d%d%d%d", &a, &b, &c, &d) != EOF) {
if (d > 2 * c || d >= b || c > 2 * d) {
printf("-1\n");
} else {
printf("%d\n%d\n%d\n", 2 * a, 2 * b, max(c, d));
}
}
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
while (scanf("%d%d%d%d", &a, &b, &c, &d) != EOF) {
if (d > 2 * c || d >= b || c > 2 * d) {
printf("-1\n");
} else {
printf("%d\n%d\n%d\n", 2 * a, 2 * b, max(c, d));
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int v[4];
int main() {
for (int i = 0; i < 4; ++i) scanf("%d", v + i);
int mn = min(v[2], v[3]);
int mx = max(v[2], v[3]);
int c = mx;
int b = max(v[1], 2 * c + 1);
int a = max(v[0], 2 * v[1] + 1);
if (mx <= 2 * mn && v[1] * 2 >= b && v[0] * 2 >= a) {
printf("%d\n%d\n%d\n", a, b, c);
} else
puts("-1");
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int v[4];
int main() {
for (int i = 0; i < 4; ++i) scanf("%d", v + i);
int mn = min(v[2], v[3]);
int mx = max(v[2], v[3]);
int c = mx;
int b = max(v[1], 2 * c + 1);
int a = max(v[0], 2 * v[1] + 1);
if (mx <= 2 * mn && v[1] * 2 >= b && v[0] * 2 >= a) {
printf("%d\n%d\n%d\n", a, b, c);
} else
puts("-1");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long i, j, k, n, y, r, z, t, tt, x, b[322567], a[334563], s, l, c[335444],
d[222222], e[222222], m;
pair<long long, long long> p[12345], q, pp;
int main() {
cin >> n >> m >> k >> x;
if (2 * k < x || 2 * x < k || x >= m) {
cout << -1;
return 0;
}
cout << 2 * n << endl << max(2 * x, m) + 1 << endl << max(k, x);
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long i, j, k, n, y, r, z, t, tt, x, b[322567], a[334563], s, l, c[335444],
d[222222], e[222222], m;
pair<long long, long long> p[12345], q, pp;
int main() {
cin >> n >> m >> k >> x;
if (2 * k < x || 2 * x < k || x >= m) {
cout << -1;
return 0;
}
cout << 2 * n << endl << max(2 * x, m) + 1 << endl << max(k, x);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long powmod(long long a, long long b) {
long long res = 1;
a %= mod;
assert(b >= 0);
for (; b; b >>= 1) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
}
return res;
}
int main() {
long long a, b, c, d;
cin >> a >> b >> c >> d;
a *= 2, b *= 2, c *= 2;
if ((d < a) && (d < b) && (d <= c)) {
if (d < c / 2)
c /= 2;
else
c = d;
if ((2 * d >= c)) {
if (2 * d < b) {
cout << a << "\n" << b << "\n" << c;
} else {
cout << -1;
}
} else {
cout << -1;
}
} else {
cout << -1;
}
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long powmod(long long a, long long b) {
long long res = 1;
a %= mod;
assert(b >= 0);
for (; b; b >>= 1) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
}
return res;
}
int main() {
long long a, b, c, d;
cin >> a >> b >> c >> d;
a *= 2, b *= 2, c *= 2;
if ((d < a) && (d < b) && (d <= c)) {
if (d < c / 2)
c /= 2;
else
c = d;
if ((2 * d >= c)) {
if (2 * d < b) {
cout << a << "\n" << b << "\n" << c;
} else {
cout << -1;
}
} else {
cout << -1;
}
} else {
cout << -1;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int vb, vs, vm, vd;
bool ok(int v1, int v2, int v3) {
if ((v3 >= max(vm, vd)) && (v3 <= 2 * vm) && (v3 <= 2 * vd) &&
(v2 >= max(vs, vd)) && (v2 <= 2 * vs) && (v2 > 2 * vd) &&
(v1 >= max(vb, vd)) && (v1 <= 2 * vb) && (v1 > 2 * vd) && (v1 > v2) &&
(v2 > v3))
return true;
return false;
}
int main() {
cin >> vb >> vs >> vm >> vd;
for (int i = 1; i <= 200; ++i) {
for (int j = 1; j <= 200; ++j) {
for (int k = 1; k <= 200; ++k) {
if (ok(i, j, k)) {
cout << i << ' ' << j << ' ' << k;
return 0;
}
}
}
}
cout << -1;
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int vb, vs, vm, vd;
bool ok(int v1, int v2, int v3) {
if ((v3 >= max(vm, vd)) && (v3 <= 2 * vm) && (v3 <= 2 * vd) &&
(v2 >= max(vs, vd)) && (v2 <= 2 * vs) && (v2 > 2 * vd) &&
(v1 >= max(vb, vd)) && (v1 <= 2 * vb) && (v1 > 2 * vd) && (v1 > v2) &&
(v2 > v3))
return true;
return false;
}
int main() {
cin >> vb >> vs >> vm >> vd;
for (int i = 1; i <= 200; ++i) {
for (int j = 1; j <= 200; ++j) {
for (int k = 1; k <= 200; ++k) {
if (ok(i, j, k)) {
cout << i << ' ' << j << ' ' << k;
return 0;
}
}
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long a[4], i, ans[4] = {0};
for (int i = 0; i < (4); i++) cin >> a[i];
for (int i = 0; i < (2); i++) ans[i] = a[i] * 2;
for (i = a[2]; i <= a[2] * 2; i++) {
if (a[3] <= i && a[3] * 2 >= i && 2 * a[3] < ans[1]) {
ans[2] = i;
for (int i = 0; i < (3); i++) cout << ans[i] << endl;
return 0;
}
}
cout << -1;
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long a[4], i, ans[4] = {0};
for (int i = 0; i < (4); i++) cin >> a[i];
for (int i = 0; i < (2); i++) ans[i] = a[i] * 2;
for (i = a[2]; i <= a[2] * 2; i++) {
if (a[3] <= i && a[3] * 2 >= i && 2 * a[3] < ans[1]) {
ans[2] = i;
for (int i = 0; i < (3); i++) cout << ans[i] << endl;
return 0;
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long V1, V2, V3, Vm;
cin >> V1 >> V2 >> V3 >> Vm;
if (Vm <= V3 && 2 * Vm >= V3 && 2 * Vm < V2 && 2 * Vm < V1) {
cout << V1 << '\n';
cout << V2 << '\n';
cout << V3 << '\n';
return 0;
}
V1 *= 2;
V2 *= 2;
V3 *= 2;
if (Vm <= V3 && 2 * Vm >= V3 && 2 * Vm < V2 && 2 * Vm < V1) {
cout << V1 << '\n';
cout << V2 << '\n';
cout << V3 << '\n';
return 0;
}
V1 /= 2;
V2 /= 2;
V3 /= 2;
V1 += V2;
V2 += V3;
if (Vm <= V3 && 2 * Vm >= V3 && 2 * Vm < V2 && 2 * Vm < V1) {
cout << V1 << '\n';
cout << V2 << '\n';
cout << V3 << '\n';
return 0;
}
cout << "-1";
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long V1, V2, V3, Vm;
cin >> V1 >> V2 >> V3 >> Vm;
if (Vm <= V3 && 2 * Vm >= V3 && 2 * Vm < V2 && 2 * Vm < V1) {
cout << V1 << '\n';
cout << V2 << '\n';
cout << V3 << '\n';
return 0;
}
V1 *= 2;
V2 *= 2;
V3 *= 2;
if (Vm <= V3 && 2 * Vm >= V3 && 2 * Vm < V2 && 2 * Vm < V1) {
cout << V1 << '\n';
cout << V2 << '\n';
cout << V3 << '\n';
return 0;
}
V1 /= 2;
V2 /= 2;
V3 /= 2;
V1 += V2;
V2 += V3;
if (Vm <= V3 && 2 * Vm >= V3 && 2 * Vm < V2 && 2 * Vm < V1) {
cout << V1 << '\n';
cout << V2 << '\n';
cout << V3 << '\n';
return 0;
}
cout << "-1";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a, b, c, k, m;
int main() {
cin >> a >> b >> c >> k;
if ((c * 2 >= k && c <= k) || (c >= k && c <= 2 * k)) {
m = min(max(k, c), c * 2);
if ((2 * b >= m && b <= m)) {
cout << -1;
return 0;
} else {
cout << a * 2 << " " << b * 2 << " ";
}
cout << m;
} else {
cout << -1;
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a, b, c, k, m;
int main() {
cin >> a >> b >> c >> k;
if ((c * 2 >= k && c <= k) || (c >= k && c <= 2 * k)) {
m = min(max(k, c), c * 2);
if ((2 * b >= m && b <= m)) {
cout << -1;
return 0;
} else {
cout << a * 2 << " " << b * 2 << " ";
}
cout << m;
} else {
cout << -1;
return 0;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int main() {
ios_base ::sync_with_stdio(0);
cin.tie(0);
int l, m, s, v;
cin >> l >> m >> s >> v;
for (int i = l; i <= 2 * l; ++i) {
for (int j = m; j <= min(2 * m, i - 1); ++j) {
for (int k = s; k <= min(2 * s, j - 1); ++k) {
if (i >= 0 && j >= 0 && k >= 0 && v <= k && 2 * v >= k &&
!(2 * v >= j) && !(2 * v >= i)) {
return cout << i << '\n' << j << '\n' << k, 0;
}
}
}
}
cout << -1;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int main() {
ios_base ::sync_with_stdio(0);
cin.tie(0);
int l, m, s, v;
cin >> l >> m >> s >> v;
for (int i = l; i <= 2 * l; ++i) {
for (int j = m; j <= min(2 * m, i - 1); ++j) {
for (int k = s; k <= min(2 * s, j - 1); ++k) {
if (i >= 0 && j >= 0 && k >= 0 && v <= k && 2 * v >= k &&
!(2 * v >= j) && !(2 * v >= i)) {
return cout << i << '\n' << j << '\n' << k, 0;
}
}
}
}
cout << -1;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
const int INF = 1e9;
const long long LINF = 1LL * INF * INF;
const int MAXN = 200007;
const double EPS = 1e-7;
int main() {
ios_base::sync_with_stdio(0);
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
for (int b1 = (0); b1 < (200); ++b1)
for (int b2 = (0); b2 < (b1); ++b2)
for (int b3 = (0); b3 < (b2); ++b3) {
if (v1 <= b1 && 2 * v1 >= b1 && v2 <= b2 && 2 * v2 >= b2 && v3 <= b3 &&
2 * v3 >= b3 && vm <= b3 && 2 * vm >= b3 && 2 * vm < b2) {
cout << b1 << endl << b2 << endl << b3 << endl;
return 0;
}
}
cout << -1;
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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
const int INF = 1e9;
const long long LINF = 1LL * INF * INF;
const int MAXN = 200007;
const double EPS = 1e-7;
int main() {
ios_base::sync_with_stdio(0);
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
for (int b1 = (0); b1 < (200); ++b1)
for (int b2 = (0); b2 < (b1); ++b2)
for (int b3 = (0); b3 < (b2); ++b3) {
if (v1 <= b1 && 2 * v1 >= b1 && v2 <= b2 && 2 * v2 >= b2 && v3 <= b3 &&
2 * v3 >= b3 && vm <= b3 && 2 * vm >= b3 && 2 * vm < b2) {
cout << b1 << endl << b2 << endl << b3 << endl;
return 0;
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
while (scanf("%d %d %d %d", &v1, &v2, &v3, &vm) == 4) {
if (vm < v2) {
if (vm > v3) {
if (2 * v3 >= vm) {
printf("%d\n%d\n%d\n", 2 * v1, 2 * v2, vm);
} else
printf("-1\n");
} else {
if (2 * vm >= v3) {
printf("%d\n%d\n%d\n", 2 * v1, 2 * v2, v3);
} else
printf("-1\n");
}
} else
printf("-1\n");
}
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
while (scanf("%d %d %d %d", &v1, &v2, &v3, &vm) == 4) {
if (vm < v2) {
if (vm > v3) {
if (2 * v3 >= vm) {
printf("%d\n%d\n%d\n", 2 * v1, 2 * v2, vm);
} else
printf("-1\n");
} else {
if (2 * vm >= v3) {
printf("%d\n%d\n%d\n", 2 * v1, 2 * v2, v3);
} else
printf("-1\n");
}
} else
printf("-1\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d;
int main() {
cin >> a >> b >> c >> d;
int tem = 0x3f3f3f3f;
if (d >= a || d >= b) {
return cout << -1, 0;
}
for (int i = c; i <= c + c; i++) {
if (i <= d + d && i >= d) {
tem = i;
break;
}
}
if (tem == 0x3f3f3f3f) {
return cout << -1, 0;
} else {
cout << a + a << '\n' << b + b << '\n' << tem << '\n';
}
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d;
int main() {
cin >> a >> b >> c >> d;
int tem = 0x3f3f3f3f;
if (d >= a || d >= b) {
return cout << -1, 0;
}
for (int i = c; i <= c + c; i++) {
if (i <= d + d && i >= d) {
tem = i;
break;
}
}
if (tem == 0x3f3f3f3f) {
return cout << -1, 0;
} else {
cout << a + a << '\n' << b + b << '\n' << tem << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline void OPEN(string s) {
freopen((s + ".in").c_str(), "r", stdin);
freopen((s + ".out").c_str(), "w", stdout);
}
int a[10];
int ans[10];
int main() {
bool flag = true;
for (int i = 1; i <= 4; i++) scanf("%d", &a[i]);
ans[3] = max(a[3], a[4]);
if (ans[3] > 2 * a[4]) flag = false;
if (ans[3] > 2 * a[3]) flag = false;
ans[2] = 2 * a[2], ans[1] = 2 * a[1];
if (ans[2] <= ans[3] * 2) flag = false;
if (!flag)
printf("-1\n");
else
printf("%d %d %d\n", ans[1], ans[2], ans[3]);
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline void OPEN(string s) {
freopen((s + ".in").c_str(), "r", stdin);
freopen((s + ".out").c_str(), "w", stdout);
}
int a[10];
int ans[10];
int main() {
bool flag = true;
for (int i = 1; i <= 4; i++) scanf("%d", &a[i]);
ans[3] = max(a[3], a[4]);
if (ans[3] > 2 * a[4]) flag = false;
if (ans[3] > 2 * a[3]) flag = false;
ans[2] = 2 * a[2], ans[1] = 2 * a[1];
if (ans[2] <= ans[3] * 2) flag = false;
if (!flag)
printf("-1\n");
else
printf("%d %d %d\n", ans[1], ans[2], ans[3]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d, x;
cin >> a >> b >> c >> d;
if (d <= 2 * c && d < b && c <= 2 * d) {
cout << a * 2 << "\n" << b * 2 << "\n";
x = (c >= d) ? c : d;
cout << x << endl;
} else
cout << "-1";
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d, x;
cin >> a >> b >> c >> d;
if (d <= 2 * c && d < b && c <= 2 * d) {
cout << a * 2 << "\n" << b * 2 << "\n";
x = (c >= d) ? c : d;
cout << x << endl;
} else
cout << "-1";
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
scanf("%d %d %d %d", &v1, &v2, &v3, &vm);
bool ok = false;
for (int c1 = max(v1, 2 * vm + 1); c1 <= 2 * v1 && !ok; c1++) {
for (int c2 = max(v2, 2 * vm + 1); c2 <= 2 * v2 && c2 < c1 && !ok; c2++) {
for (int c3 = max(v3, vm); c3 <= 2 * v3 && c3 <= 2 * vm && c3 < c2 && !ok;
c3++) {
ok = true;
printf("%d\n%d\n%d\n", c1, c2, c3);
}
}
}
if (!ok) printf("-1\n");
return 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
scanf("%d %d %d %d", &v1, &v2, &v3, &vm);
bool ok = false;
for (int c1 = max(v1, 2 * vm + 1); c1 <= 2 * v1 && !ok; c1++) {
for (int c2 = max(v2, 2 * vm + 1); c2 <= 2 * v2 && c2 < c1 && !ok; c2++) {
for (int c3 = max(v3, vm); c3 <= 2 * v3 && c3 <= 2 * vm && c3 < c2 && !ok;
c3++) {
ok = true;
printf("%d\n%d\n%d\n", c1, c2, c3);
}
}
}
if (!ok) printf("-1\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, v4;
int main() {
cin >> v1 >> v2 >> v3 >> v4;
int a[] = {v1 * 2, v2 * 2, v3 * 2};
int ans[3];
if (v4 > a[2] || v4 * 2 < v3)
cout << "-1";
else {
ans[2] = max(v4, v3);
ans[1] = max(2 * v4 + 1, v2);
ans[0] = max(2 * v4 + 2, v1);
if (ans[1] > a[1] || ans[0] > a[0])
cout << "-1";
else
cout << ans[0] << "\n" << ans[1] << "\n" << ans[2];
}
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, v4;
int main() {
cin >> v1 >> v2 >> v3 >> v4;
int a[] = {v1 * 2, v2 * 2, v3 * 2};
int ans[3];
if (v4 > a[2] || v4 * 2 < v3)
cout << "-1";
else {
ans[2] = max(v4, v3);
ans[1] = max(2 * v4 + 1, v2);
ans[0] = max(2 * v4 + 2, v1);
if (ans[1] > a[1] || ans[0] > a[0])
cout << "-1";
else
cout << ans[0] << "\n" << ans[1] << "\n" << ans[2];
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
float inf = std::numeric_limits<double>::infinity();
long long int INF = std::numeric_limits<long long int>::max();
int main() {
int a, b, c, d;
scanf("%d", &a);
scanf("%d", &b);
scanf("%d", &c);
scanf("%d", &d);
int x, y, z;
z = max(c, d);
if (z > 2 * c || z > 2 * d) return 0 * puts("-1");
y = max(b, max(z, 2 * d) + 1);
if (y > 2 * b) return 0 * puts("-1");
x = max(a, y + 1);
if (x > 2 * a) return 0 * puts("-1");
printf("%d %d %d\n", x, y, z);
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cerr.write(names, comma - names) << " : " << arg1 << " | ";
__f(comma + 1, args...);
}
float inf = std::numeric_limits<double>::infinity();
long long int INF = std::numeric_limits<long long int>::max();
int main() {
int a, b, c, d;
scanf("%d", &a);
scanf("%d", &b);
scanf("%d", &c);
scanf("%d", &d);
int x, y, z;
z = max(c, d);
if (z > 2 * c || z > 2 * d) return 0 * puts("-1");
y = max(b, max(z, 2 * d) + 1);
if (y > 2 * b) return 0 * puts("-1");
x = max(a, y + 1);
if (x > 2 * a) return 0 * puts("-1");
printf("%d %d %d\n", x, y, z);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int m_size = 0, f_size = 0, s_size = 0, masha_size = 0;
cin >> f_size >> m_size >> s_size >> masha_size;
int f_car = -1, m_car = -1, s_car = -1;
for (int i = 2 * f_size; i >= f_size; i--) {
if (i > 2 * masha_size) {
f_car = i;
break;
}
}
for (int i = 2 * m_size; i >= m_size; i--) {
if (i > 2 * masha_size && i < f_car) {
m_car = i;
break;
}
}
for (int i = 2 * s_size; i >= s_size; i--) {
if (i >= masha_size && i <= 2 * masha_size && i < m_car) {
s_car = i;
break;
}
}
if (f_car == -1 || m_car == -1 || s_car == -1)
cout << -1;
else {
cout << f_car << endl << m_car << endl << s_car << endl;
}
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int m_size = 0, f_size = 0, s_size = 0, masha_size = 0;
cin >> f_size >> m_size >> s_size >> masha_size;
int f_car = -1, m_car = -1, s_car = -1;
for (int i = 2 * f_size; i >= f_size; i--) {
if (i > 2 * masha_size) {
f_car = i;
break;
}
}
for (int i = 2 * m_size; i >= m_size; i--) {
if (i > 2 * masha_size && i < f_car) {
m_car = i;
break;
}
}
for (int i = 2 * s_size; i >= s_size; i--) {
if (i >= masha_size && i <= 2 * masha_size && i < m_car) {
s_car = i;
break;
}
}
if (f_car == -1 || m_car == -1 || s_car == -1)
cout << -1;
else {
cout << f_car << endl << m_car << endl << s_car << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
ifstream in("input.txt");
ofstream out("output.txt");
const int MAXN = 1e5 + 7;
const long long MAXL = 1e18;
const int N = 1e7;
const double eps = 1e-11;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int mini = 1; mini <= 200; ++mini) {
for (int sr = mini + 1; sr <= 200; ++sr) {
for (int maxi = sr + 1; maxi <= 200; ++maxi) {
if (a <= maxi && a * 2 >= maxi && b <= sr && b * 2 >= sr && c <= mini &&
c * 2 >= mini && d <= maxi && d * 2 < maxi && d <= sr &&
d * 2 < sr && d <= mini && d * 2 >= mini) {
cout << maxi << "\n" << sr << "\n" << mini << "\n";
return 0;
}
}
}
}
cout << -1 << "\n";
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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
ifstream in("input.txt");
ofstream out("output.txt");
const int MAXN = 1e5 + 7;
const long long MAXL = 1e18;
const int N = 1e7;
const double eps = 1e-11;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int mini = 1; mini <= 200; ++mini) {
for (int sr = mini + 1; sr <= 200; ++sr) {
for (int maxi = sr + 1; maxi <= 200; ++maxi) {
if (a <= maxi && a * 2 >= maxi && b <= sr && b * 2 >= sr && c <= mini &&
c * 2 >= mini && d <= maxi && d * 2 < maxi && d <= sr &&
d * 2 < sr && d <= mini && d * 2 >= mini) {
cout << maxi << "\n" << sr << "\n" << mini << "\n";
return 0;
}
}
}
}
cout << -1 << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
for (int i = 1; i <= 200; i++) {
for (int j = i + 1; j <= 200; j++) {
for (int k = j + 1; k <= 200; k++) {
if ((v1 <= k) && (v2 <= j) && (v3 <= i) && (vm <= k) && (vm <= j) &&
(vm <= i) && (2 * v1 >= k) && (2 * v2 >= j) && (2 * v3 >= i) &&
(2 * vm >= i) && (2 * vm < j) && (2 * vm < k)) {
cout << k << "\n" << j << "\n" << i << "\n";
return 0;
}
}
}
}
cout << -1 << "\n";
}
| ### 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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
for (int i = 1; i <= 200; i++) {
for (int j = i + 1; j <= 200; j++) {
for (int k = j + 1; k <= 200; k++) {
if ((v1 <= k) && (v2 <= j) && (v3 <= i) && (vm <= k) && (vm <= j) &&
(vm <= i) && (2 * v1 >= k) && (2 * v2 >= j) && (2 * v3 >= i) &&
(2 * vm >= i) && (2 * vm < j) && (2 * vm < k)) {
cout << k << "\n" << j << "\n" << i << "\n";
return 0;
}
}
}
}
cout << -1 << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(false), cin.tie(0);
;
int fb, mb, sb, m;
cin >> fb >> mb >> sb >> m;
bool possible = false;
for (int i = 1; i < 201; i++) {
for (int j = i + 1; j < 201; j++) {
for (int k = j + 1; k < 201; k++) {
if ((i >= m && i <= 2 * m) && !(j >= m && j <= 2 * m) &&
!(k >= m && k <= 2 * m)) {
if ((i >= sb && i <= 2 * sb) && (j >= mb && j <= 2 * mb) &&
(k >= fb && k <= 2 * fb)) {
cout << k << "\n" << j << "\n" << i << "\n";
possible = true;
goto looped;
}
}
}
}
}
looped:
if (!possible) {
cout << -1 << '\n';
}
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(false), cin.tie(0);
;
int fb, mb, sb, m;
cin >> fb >> mb >> sb >> m;
bool possible = false;
for (int i = 1; i < 201; i++) {
for (int j = i + 1; j < 201; j++) {
for (int k = j + 1; k < 201; k++) {
if ((i >= m && i <= 2 * m) && !(j >= m && j <= 2 * m) &&
!(k >= m && k <= 2 * m)) {
if ((i >= sb && i <= 2 * sb) && (j >= mb && j <= 2 * mb) &&
(k >= fb && k <= 2 * fb)) {
cout << k << "\n" << j << "\n" << i << "\n";
possible = true;
goto looped;
}
}
}
}
}
looped:
if (!possible) {
cout << -1 << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dad, mum, son, masha;
bool ccheck(int tx, int ty, int tz) {
if (dad > tx) return 0;
if (2 * dad < tx) return 0;
if (mum > ty) return 0;
if (2 * mum < ty) return 0;
if (son > tz) return 0;
if (2 * son < tz) return 0;
if (masha > tz) return 0;
if (masha * 2 < tz) return 0;
if (masha * 2 >= ty) return 0;
return 1;
}
int main() {
ios::sync_with_stdio(false);
cin >> dad >> mum >> son >> masha;
for (int i = 1; i <= 250; i++) {
for (int j = 1; j < i; j++) {
for (int k = 1; k < j; k++) {
if (ccheck(i, j, k)) {
cout << i << endl << j << endl << k << endl;
exit(0);
}
}
}
}
cout << -1;
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dad, mum, son, masha;
bool ccheck(int tx, int ty, int tz) {
if (dad > tx) return 0;
if (2 * dad < tx) return 0;
if (mum > ty) return 0;
if (2 * mum < ty) return 0;
if (son > tz) return 0;
if (2 * son < tz) return 0;
if (masha > tz) return 0;
if (masha * 2 < tz) return 0;
if (masha * 2 >= ty) return 0;
return 1;
}
int main() {
ios::sync_with_stdio(false);
cin >> dad >> mum >> son >> masha;
for (int i = 1; i <= 250; i++) {
for (int j = 1; j < i; j++) {
for (int k = 1; k < j; k++) {
if (ccheck(i, j, k)) {
cout << i << endl << j << endl << k << endl;
exit(0);
}
}
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int a, b, c, d;
cin >> c >> b >> a >> d;
int x, y, z;
x = -1, y = -1, z = -1;
int j = 0;
int arr[100], brr[100], crr[100];
for (int i = a; i <= 2 * a; i++) {
if (i <= 2 * d && i >= d) {
x = i;
arr[j] = i;
j++;
}
}
int sz1 = j;
j = 0;
for (int i = b; i <= 2 * b; i++) {
if (i > 2 * d && i >= d) {
y = i;
brr[j] = i;
j++;
}
}
int sz2 = j;
j = 0;
for (int i = c; i <= 2 * c; i++) {
if (i > 2 * d && i >= d) {
z = i;
crr[j] = i;
j++;
}
}
int sz3 = j;
int f = 0;
for (int i = sz1 - 1; i >= 0; i--) {
for (j = sz2 - 1; j >= 0; j--) {
for (int k = sz3 - 1; k >= 0; k--) {
if (crr[k] > brr[j] && brr[j] > arr[i]) {
x = arr[i];
y = brr[j];
z = crr[k];
f = 1;
break;
}
}
if (f == 1) break;
}
if (f == 1) break;
}
if (x == -1 || y == -1 || z == -1 || f == 0) {
cout << "-1";
} else {
cout << z << endl << y << endl << x << endl;
}
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int a, b, c, d;
cin >> c >> b >> a >> d;
int x, y, z;
x = -1, y = -1, z = -1;
int j = 0;
int arr[100], brr[100], crr[100];
for (int i = a; i <= 2 * a; i++) {
if (i <= 2 * d && i >= d) {
x = i;
arr[j] = i;
j++;
}
}
int sz1 = j;
j = 0;
for (int i = b; i <= 2 * b; i++) {
if (i > 2 * d && i >= d) {
y = i;
brr[j] = i;
j++;
}
}
int sz2 = j;
j = 0;
for (int i = c; i <= 2 * c; i++) {
if (i > 2 * d && i >= d) {
z = i;
crr[j] = i;
j++;
}
}
int sz3 = j;
int f = 0;
for (int i = sz1 - 1; i >= 0; i--) {
for (j = sz2 - 1; j >= 0; j--) {
for (int k = sz3 - 1; k >= 0; k--) {
if (crr[k] > brr[j] && brr[j] > arr[i]) {
x = arr[i];
y = brr[j];
z = crr[k];
f = 1;
break;
}
}
if (f == 1) break;
}
if (f == 1) break;
}
if (x == -1 || y == -1 || z == -1 || f == 0) {
cout << "-1";
} else {
cout << z << endl << y << endl << x << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
scanf("%d%d%d%d", &v1, &v2, &v3, &v4);
if (2 * v3 < v4 || v1 <= v4 || v2 <= v4 || 2 * v4 < v3) {
printf("-1\n");
} else {
printf("%d\n%d\n", 2 * v1, 2 * v2);
if (v4 >= v3)
printf("%d\n", 2 * v3);
else {
printf("%d\n", v3);
}
}
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
scanf("%d%d%d%d", &v1, &v2, &v3, &v4);
if (2 * v3 < v4 || v1 <= v4 || v2 <= v4 || 2 * v4 < v3) {
printf("-1\n");
} else {
printf("%d\n%d\n", 2 * v1, 2 * v2);
if (v4 >= v3)
printf("%d\n", 2 * v3);
else {
printf("%d\n", v3);
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
int main() {
scanf("%d %d %d %d", &v1, &v2, &v3, &vm);
for (int s1 = max(2 * vm + 1, v1); s1 <= 2 * v1; s1++)
for (int s2 = max(2 * vm + 1, v2); s2 < s1 && s2 <= 2 * v2; s2++)
for (int s3 = max(vm, v3); s3 < s2 && s3 <= 2 * v3 && s3 <= 2 * vm;
s3++) {
printf("%d\n%d\n%d\n", s1, s2, s3);
return 0;
}
printf("-1\n");
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
int main() {
scanf("%d %d %d %d", &v1, &v2, &v3, &vm);
for (int s1 = max(2 * vm + 1, v1); s1 <= 2 * v1; s1++)
for (int s2 = max(2 * vm + 1, v2); s2 < s1 && s2 <= 2 * v2; s2++)
for (int s3 = max(vm, v3); s3 < s2 && s3 <= 2 * v3 && s3 <= 2 * vm;
s3++) {
printf("%d\n%d\n%d\n", s1, s2, s3);
return 0;
}
printf("-1\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
int main() {
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
if (vm > 2 * v3 || vm < v3 / 2 + (v3 & 1) || vm >= v2) cout << -1, exit(0);
cout << 2 * v1 << endl << 2 * v2 << endl << max(vm, v3);
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
int main() {
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
if (vm > 2 * v3 || vm < v3 / 2 + (v3 & 1) || vm >= v2) cout << -1, exit(0);
cout << 2 * v1 << endl << 2 * v2 << endl << max(vm, v3);
return 0;
}
``` |
#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);
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#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);
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d, max = 0;
scanf("%d %d %d %d", &a, &b, &c, &d);
if (d < a && d < b && 2 * d >= c && d <= 2 * c) {
if (c >= d)
max = c;
else
max = d;
printf("%d %d %d", 2 * a, 2 * b, max);
} else {
printf("%d", -1);
}
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d, max = 0;
scanf("%d %d %d %d", &a, &b, &c, &d);
if (d < a && d < b && 2 * d >= c && d <= 2 * c) {
if (c >= d)
max = c;
else
max = d;
printf("%d %d %d", 2 * a, 2 * b, max);
} else {
printf("%d", -1);
}
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int N = 2e5 + 7;
const int M = 1e6 + 7;
signed main() {
ios::sync_with_stdio(false);
int fa, ma, son, sb;
cin >> fa >> ma >> son >> sb;
int lim[7];
for (int c3 = max(son, sb); c3 <= min(2 * son, 2 * sb); c3++)
for (int c2 = max(ma, max(c3 + 1, 2 * sb + 1)); c2 <= 2 * ma; c2++)
for (int c1 = max(fa, max(c2 + 1, 2 * sb + 1)); c3 <= 2 * fa; c3++) {
cout << c1 << "\n" << c2 << "\n" << c3 << "\n";
return 0;
}
cout << "-1\n";
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int N = 2e5 + 7;
const int M = 1e6 + 7;
signed main() {
ios::sync_with_stdio(false);
int fa, ma, son, sb;
cin >> fa >> ma >> son >> sb;
int lim[7];
for (int c3 = max(son, sb); c3 <= min(2 * son, 2 * sb); c3++)
for (int c2 = max(ma, max(c3 + 1, 2 * sb + 1)); c2 <= 2 * ma; c2++)
for (int c1 = max(fa, max(c2 + 1, 2 * sb + 1)); c3 <= 2 * fa; c3++) {
cout << c1 << "\n" << c2 << "\n" << c3 << "\n";
return 0;
}
cout << "-1\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
if (v3 > 2 * vm || vm > 2 * v3 || vm >= v2)
cout << -1;
else
cout << 2 * v1 << endl << 2 * v2 << endl << max(v3, vm);
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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
if (v3 > 2 * vm || vm > 2 * v3 || vm >= v2)
cout << -1;
else
cout << 2 * v1 << endl << 2 * v2 << endl << max(v3, vm);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int v[4];
int main() {
for (int i = 0; i < 4; ++i) scanf("%d", v + i);
int mn = min(v[2], v[3]);
int mx = max(v[2], v[3]);
int c = mx;
int b = max(v[1], 2 * c + 1);
int a = max(v[0], b + 1);
if (mx <= 2 * mn && v[1] * 2 >= b && v[0] * 2 >= a) {
printf("%d\n%d\n%d\n", a, b, c);
} else
puts("-1");
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int v[4];
int main() {
for (int i = 0; i < 4; ++i) scanf("%d", v + i);
int mn = min(v[2], v[3]);
int mx = max(v[2], v[3]);
int c = mx;
int b = max(v[1], 2 * c + 1);
int a = max(v[0], b + 1);
if (mx <= 2 * mn && v[1] * 2 >= b && v[0] * 2 >= a) {
printf("%d\n%d\n%d\n", a, b, c);
} else
puts("-1");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4, ans;
scanf("%d %d %d %d", &v1, &v2, &v3, &v4);
if (v4 > v3 * 2 || v3 > v4 * 2 || v4 >= v1 || v4 >= v2) {
printf("-1");
return 0;
}
ans = max(v3, v4);
printf("%d %d %d", v1 * 2, v2 * 2, ans);
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4, ans;
scanf("%d %d %d %d", &v1, &v2, &v3, &v4);
if (v4 > v3 * 2 || v3 > v4 * 2 || v4 >= v1 || v4 >= v2) {
printf("-1");
return 0;
}
ans = max(v3, v4);
printf("%d %d %d", v1 * 2, v2 * 2, ans);
}
``` |
#include <bits/stdc++.h>
using namespace std;
int V1, V2, V3, T1, T2, T3, Vm;
int main() {
cin >> V1 >> V2 >> V3 >> Vm;
if (V3 > 2 * Vm || Vm > 2 * V3) {
cout << -1;
return 0;
}
T3 = max(Vm, V3);
if (T3 >= V2) {
cout << -1;
return 0;
}
T2 = min(2 * V2, max(V2, 2 * T3 + 1));
T1 = min(2 * V1, max(V1, T2 + 1));
cout << T1 << endl << T2 << endl << T3;
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int V1, V2, V3, T1, T2, T3, Vm;
int main() {
cin >> V1 >> V2 >> V3 >> Vm;
if (V3 > 2 * Vm || Vm > 2 * V3) {
cout << -1;
return 0;
}
T3 = max(Vm, V3);
if (T3 >= V2) {
cout << -1;
return 0;
}
T2 = min(2 * V2, max(V2, 2 * T3 + 1));
T1 = min(2 * V1, max(V1, T2 + 1));
cout << T1 << endl << T2 << endl << T3;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long int INFL = 0x3f3f3f3f3f3f3f3f;
const double pi = atan(1.0) * 4.0;
const int N = 0;
int main(void) {
ios_base::sync_with_stdio(false);
int a, b, c, d;
cin >> a >> b >> c >> d;
int ca = 2 * a;
int cb = 2 * b;
int cc = min(2 * c, 2 * d);
if (cc >= c && 2 * c >= cc && cc >= d && 2 * d >= cc && 2 * d < cb) {
cout << ca << endl;
cout << cb << endl;
cout << cc << endl;
} else
cout << -1 << endl;
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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long int INFL = 0x3f3f3f3f3f3f3f3f;
const double pi = atan(1.0) * 4.0;
const int N = 0;
int main(void) {
ios_base::sync_with_stdio(false);
int a, b, c, d;
cin >> a >> b >> c >> d;
int ca = 2 * a;
int cb = 2 * b;
int cc = min(2 * c, 2 * d);
if (cc >= c && 2 * c >= cc && cc >= d && 2 * d >= cc && 2 * d < cb) {
cout << ca << endl;
cout << cb << endl;
cout << cc << endl;
} else
cout << -1 << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
if (d > 2 * c || c > 2 * d || d >= b)
printf("-1");
else
printf("%d\n%d\n%d", 2 * a, 2 * b, max(c, d));
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
if (d > 2 * c || c > 2 * d || d >= b)
printf("-1");
else
printf("%d\n%d\n%d", 2 * a, 2 * b, max(c, d));
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
while (~scanf("%d%d%d%d", &v1, &v2, &v3, &vm)) {
if (2 * vm >= v3 && vm <= 2 * v3 && vm < v2) {
int a1 = 2 * v1, a2 = 2 * v2, a3 = (vm <= v3) ? v3 : vm;
printf("%d\n%d\n%d\n", a1, a2, a3);
} else
printf("-1\n");
}
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, vm;
while (~scanf("%d%d%d%d", &v1, &v2, &v3, &vm)) {
if (2 * vm >= v3 && vm <= 2 * v3 && vm < v2) {
int a1 = 2 * v1, a2 = 2 * v2, a3 = (vm <= v3) ? v3 : vm;
printf("%d\n%d\n%d\n", a1, a2, a3);
} else
printf("-1\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if (d * 2 >= c && c * 2 >= d && d < b) {
int x = max(c, d);
int y = max(x * 2 + 1, b);
int z = max(y + 1, a);
cout << z << "\n" << y << "\n" << x << endl;
} else
cout << "-1" << 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if (d * 2 >= c && c * 2 >= d && d < b) {
int x = max(c, d);
int y = max(x * 2 + 1, b);
int z = max(y + 1, a);
cout << z << "\n" << y << "\n" << x << endl;
} else
cout << "-1" << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9;
const long long inf64 = 1e18;
const long long MOD = inf + 7;
const long double PI = acos(-1.0);
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
;
long long a, b, c, d;
cin >> a >> b >> c >> d;
for (long long i = c; i <= 2 * c; i++) {
if (d <= i && 2 * d >= i && b > i) {
cout << 2 * a << endl << 2 * b << endl << i;
return 0;
}
}
cout << -1;
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9;
const long long inf64 = 1e18;
const long long MOD = inf + 7;
const long double PI = acos(-1.0);
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
;
long long a, b, c, d;
cin >> a >> b >> c >> d;
for (long long i = c; i <= 2 * c; i++) {
if (d <= i && 2 * d >= i && b > i) {
cout << 2 * a << endl << 2 * b << endl << i;
return 0;
}
}
cout << -1;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, k, v[4], m = 4;
for (i = 0; i < m; i++) cin >> v[i];
int masha = v[3];
for (k = v[0]; k <= 2 * v[0]; k++)
for (j = v[1]; j <= 2 * v[1]; j++)
for (i = masha; i <= 2 * v[2]; i++)
if (i <= 2 * masha && j > 2 * masha && k > 2 * masha && k > j &&
j > i && i >= v[2]) {
cout << k << ' ' << j << ' ' << i << endl;
return 0;
}
cout << -1 << endl;
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, k, v[4], m = 4;
for (i = 0; i < m; i++) cin >> v[i];
int masha = v[3];
for (k = v[0]; k <= 2 * v[0]; k++)
for (j = v[1]; j <= 2 * v[1]; j++)
for (i = masha; i <= 2 * v[2]; i++)
if (i <= 2 * masha && j > 2 * masha && k > 2 * masha && k > j &&
j > i && i >= v[2]) {
cout << k << ' ' << j << ' ' << i << endl;
return 0;
}
cout << -1 << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void build() {}
const int mod = 1000000007, N = 200005;
const long long inf = 1e18;
void preprocess() { return; }
int main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
build();
preprocess();
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
int c1, c2, c3;
int flag = 1;
if (v4 > 2 * v3) flag = 0;
c3 = max(v4, v3);
if (c3 > 2 * v4) flag = 0;
if (2 * v2 <= 2 * v4) flag = 0;
c2 = 2 * v2;
c1 = 2 * v1;
if (flag == 0)
cout << -1;
else
cout << c1 << " " << c2 << " " << c3;
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void build() {}
const int mod = 1000000007, N = 200005;
const long long inf = 1e18;
void preprocess() { return; }
int main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
build();
preprocess();
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
int c1, c2, c3;
int flag = 1;
if (v4 > 2 * v3) flag = 0;
c3 = max(v4, v3);
if (c3 > 2 * v4) flag = 0;
if (2 * v2 <= 2 * v4) flag = 0;
c2 = 2 * v2;
c1 = 2 * v1;
if (flag == 0)
cout << -1;
else
cout << c1 << " " << c2 << " " << c3;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
bool flag = 0;
for (int i = a; i <= 2 * a; i++) {
for (int j = b; j <= 2 * b; j++) {
for (int k = c; k <= 2 * c; k++) {
if (i > j && j > k && k >= d && k <= 2 * d && j > 2 * d) {
cout << i << endl << j << endl << k << endl;
return 0;
}
}
}
}
cout << -1 << endl;
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
bool flag = 0;
for (int i = a; i <= 2 * a; i++) {
for (int j = b; j <= 2 * b; j++) {
for (int k = c; k <= 2 * c; k++) {
if (i > j && j > k && k >= d && k <= 2 * d && j > 2 * d) {
cout << i << endl << j << endl << k << endl;
return 0;
}
}
}
}
cout << -1 << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
bool judge(int a, int b) { return a <= b && a * 2 >= b; }
int main(void) {
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
for (int i = 1; i <= 300; ++i) {
for (int j = 1; j < i; ++j) {
for (int k = 1; k < j; ++k) {
if (judge(v1, i) && judge(v2, j) && judge(v3, k) &&
(v4 <= k && (v4 * 2 >= k) && (v4 * 2 < j))) {
cout << i << endl << j << endl << k;
return 0;
}
}
}
}
cout << -1;
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool judge(int a, int b) { return a <= b && a * 2 >= b; }
int main(void) {
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
for (int i = 1; i <= 300; ++i) {
for (int j = 1; j < i; ++j) {
for (int k = 1; k < j; ++k) {
if (judge(v1, i) && judge(v2, j) && judge(v3, k) &&
(v4 <= k && (v4 * 2 >= k) && (v4 * 2 < j))) {
cout << i << endl << j << endl << k;
return 0;
}
}
}
}
cout << -1;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
int l[5], r[5];
int main() {
while (~scanf("%d%d%d%d", &v1, &v2, &v3, &vm)) {
l[1] = v1;
r[1] = 2 * v1;
l[2] = v2;
r[2] = 2 * v2;
l[3] = v3;
r[3] = 2 * v3;
int a1 = -1, a2 = -1, a3 = -1;
for (int i = r[1]; i >= l[1] && a1 == -1; i--) {
if (vm <= i && 2 * vm < i) a1 = i;
}
for (int i = r[2]; i >= l[2] && a2 == -1; i--) {
if (vm <= i && 2 * vm < i) a2 = i;
}
for (int i = l[3]; i <= r[3] && a3 == -1; i++) {
if (vm <= i && 2 * vm >= i) a3 = i;
}
if (a1 != -1 && a2 != -1 && a3 != -1)
printf("%d\n%d\n%d\n", a1, a2, a3);
else
puts("-1");
}
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
int l[5], r[5];
int main() {
while (~scanf("%d%d%d%d", &v1, &v2, &v3, &vm)) {
l[1] = v1;
r[1] = 2 * v1;
l[2] = v2;
r[2] = 2 * v2;
l[3] = v3;
r[3] = 2 * v3;
int a1 = -1, a2 = -1, a3 = -1;
for (int i = r[1]; i >= l[1] && a1 == -1; i--) {
if (vm <= i && 2 * vm < i) a1 = i;
}
for (int i = r[2]; i >= l[2] && a2 == -1; i--) {
if (vm <= i && 2 * vm < i) a2 = i;
}
for (int i = l[3]; i <= r[3] && a3 == -1; i++) {
if (vm <= i && 2 * vm >= i) a3 = i;
}
if (a1 != -1 && a2 != -1 && a3 != -1)
printf("%d\n%d\n%d\n", a1, a2, a3);
else
puts("-1");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void readi(int &x) {
int v = 0, f = 1;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-')
f = -1;
else
v = v * 10 + c - '0';
while (isdigit(c = getchar())) v = v * 10 + c - '0';
x = v * f;
}
void readll(long long &x) {
long long v = 0ll, f = 1ll;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-')
f = -1;
else
v = v * 10 + c - '0';
while (isdigit(c = getchar())) v = v * 10 + c - '0';
x = v * f;
}
void readc(char &x) {
char c;
while ((c = getchar()) == ' ')
;
x = c;
}
void writes(string s) { puts(s.c_str()); }
void writeln() { writes(""); }
void writei(int x) {
if (!x) putchar('0');
char a[25];
int top = 0;
while (x) {
a[++top] = (x % 10) + '0';
x /= 10;
}
while (top) {
putchar(a[top]);
top--;
}
}
void writell(long long x) {
if (!x) putchar('0');
char a[25];
int top = 0;
while (x) {
a[++top] = (x % 10) + '0';
x /= 10;
}
while (top) {
putchar(a[top]);
top--;
}
}
long long qp(long long x, long long y) {
if (y == 0) return 1;
if (y == 1) return x;
long long z = qp(x, y / 2);
z = z * z % 1000000007;
if (y & 1) z = z * x % 1000000007;
return z;
}
int a, b, c, d, i, j, k;
int main() {
cin >> a >> b >> c >> d;
for (i = 1; i <= 200; i++) {
for (j = 1; j < i; j++) {
for (k = 1; k < j; k++) {
if (a <= i && a * 2 >= i && b <= j && 2 * b >= j && c <= k &&
2 * c >= k && d <= k && 2 * d >= k && 2 * d < j) {
cout << i << endl << j << endl << k << endl;
return 0;
}
}
}
}
cout << -1;
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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void readi(int &x) {
int v = 0, f = 1;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-')
f = -1;
else
v = v * 10 + c - '0';
while (isdigit(c = getchar())) v = v * 10 + c - '0';
x = v * f;
}
void readll(long long &x) {
long long v = 0ll, f = 1ll;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-')
f = -1;
else
v = v * 10 + c - '0';
while (isdigit(c = getchar())) v = v * 10 + c - '0';
x = v * f;
}
void readc(char &x) {
char c;
while ((c = getchar()) == ' ')
;
x = c;
}
void writes(string s) { puts(s.c_str()); }
void writeln() { writes(""); }
void writei(int x) {
if (!x) putchar('0');
char a[25];
int top = 0;
while (x) {
a[++top] = (x % 10) + '0';
x /= 10;
}
while (top) {
putchar(a[top]);
top--;
}
}
void writell(long long x) {
if (!x) putchar('0');
char a[25];
int top = 0;
while (x) {
a[++top] = (x % 10) + '0';
x /= 10;
}
while (top) {
putchar(a[top]);
top--;
}
}
long long qp(long long x, long long y) {
if (y == 0) return 1;
if (y == 1) return x;
long long z = qp(x, y / 2);
z = z * z % 1000000007;
if (y & 1) z = z * x % 1000000007;
return z;
}
int a, b, c, d, i, j, k;
int main() {
cin >> a >> b >> c >> d;
for (i = 1; i <= 200; i++) {
for (j = 1; j < i; j++) {
for (k = 1; k < j; k++) {
if (a <= i && a * 2 >= i && b <= j && 2 * b >= j && c <= k &&
2 * c >= k && d <= k && 2 * d >= k && 2 * d < j) {
cout << i << endl << j << endl << k << endl;
return 0;
}
}
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int a, b, c, d;
cin >> a >> b >> c >> d;
a *= 2;
b *= 2;
if (2 * d >= a || 2 * d >= b || 2 * c < d || 2 * d < c) {
cout << -1 << endl;
return 0;
} else {
cout << a << endl;
cout << b << endl;
long long int temp = min(c, d);
cout << 2 * temp << endl;
}
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int a, b, c, d;
cin >> a >> b >> c >> d;
a *= 2;
b *= 2;
if (2 * d >= a || 2 * d >= b || 2 * c < d || 2 * d < c) {
cout << -1 << endl;
return 0;
} else {
cout << a << endl;
cout << b << endl;
long long int temp = min(c, d);
cout << 2 * temp << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
long long inf = 1LL << 60;
long long mod = 1000000007;
double pi = acos(-1.0);
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
vector<int> ans;
for (int i = v3; i <= 2 * v3; i++) {
if (i >= vm && 2 * vm >= i) {
ans.push_back(i);
break;
}
}
if (ans.size() == 0) {
cout << -1;
return 0;
}
for (int i = max(ans[0] + 1, v2); i <= 2 * v2; i++) {
if (i >= vm && 2 * vm < i) {
ans.push_back(i);
break;
}
}
if (ans.size() == 1) {
cout << -1;
return 0;
}
for (int i = max(ans[1] + 1, v1); i <= 2 * v1; i++) {
if (i >= vm && 2 * vm < i) {
ans.push_back(i);
break;
}
}
if (ans.size() == 3) {
cout << ans[2] << endl;
cout << ans[1] << endl;
cout << ans[0];
return 0;
}
cout << -1;
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
long long inf = 1LL << 60;
long long mod = 1000000007;
double pi = acos(-1.0);
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int v1, v2, v3, vm;
cin >> v1 >> v2 >> v3 >> vm;
vector<int> ans;
for (int i = v3; i <= 2 * v3; i++) {
if (i >= vm && 2 * vm >= i) {
ans.push_back(i);
break;
}
}
if (ans.size() == 0) {
cout << -1;
return 0;
}
for (int i = max(ans[0] + 1, v2); i <= 2 * v2; i++) {
if (i >= vm && 2 * vm < i) {
ans.push_back(i);
break;
}
}
if (ans.size() == 1) {
cout << -1;
return 0;
}
for (int i = max(ans[1] + 1, v1); i <= 2 * v1; i++) {
if (i >= vm && 2 * vm < i) {
ans.push_back(i);
break;
}
}
if (ans.size() == 3) {
cout << ans[2] << endl;
cout << ans[1] << endl;
cout << ans[0];
return 0;
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j <= 200; j++) {
for (int k = 1; k <= 200; k++) {
if (v1 <= i && v2 <= j && v3 <= k && 2 * v1 >= i && 2 * v2 >= j &&
2 * v3 >= k && v4 <= i && v4 <= j && v4 <= k && 2 * v4 >= k &&
2 * v4 < i && 2 * v4 < j && i > j && j > k) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return 0;
}
}
}
}
cout << -1 << endl;
return 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j <= 200; j++) {
for (int k = 1; k <= 200; k++) {
if (v1 <= i && v2 <= j && v3 <= k && 2 * v1 >= i && 2 * v2 >= j &&
2 * v3 >= k && v4 <= i && v4 <= j && v4 <= k && 2 * v4 >= k &&
2 * v4 < i && 2 * v4 < j && i > j && j > k) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return 0;
}
}
}
}
cout << -1 << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int V1, V2, V3, Vm;
int main() {
ios::sync_with_stdio(0);
cin.tie(NULL);
cin >> V1 >> V2 >> V3 >> Vm;
for (int i = (V1); i <= (2 * V1); i++)
for (int j = (V2); j <= (2 * V2); j++)
for (int k = (V3); k <= (2 * V3); k++) {
if (i > j && j > k && Vm <= k && k <= 2 * Vm && 2 * Vm < i &&
2 * Vm < j) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return 0;
}
}
cout << -1;
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int V1, V2, V3, Vm;
int main() {
ios::sync_with_stdio(0);
cin.tie(NULL);
cin >> V1 >> V2 >> V3 >> Vm;
for (int i = (V1); i <= (2 * V1); i++)
for (int j = (V2); j <= (2 * V2); j++)
for (int k = (V3); k <= (2 * V3); k++) {
if (i > j && j > k && Vm <= k && k <= 2 * Vm && 2 * Vm < i &&
2 * Vm < j) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return 0;
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, sg, i, j, k;
cin >> v1 >> v2 >> v3 >> sg;
for (i = 2 * v1; i; i--)
for (j = min(2 * v2, i - 1); j; j--)
for (k = i - 1; k; k--)
if (max(max(sg, v1), 2 * sg + 1) <= i &&
max(max(sg, v2), 2 * sg + 1) <= j && max(sg, v3) <= k &&
k <= min(2 * sg, 2 * v3)) {
cout << i << endl << j << endl << k << endl;
return 0;
}
cout << -1 << endl;
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, sg, i, j, k;
cin >> v1 >> v2 >> v3 >> sg;
for (i = 2 * v1; i; i--)
for (j = min(2 * v2, i - 1); j; j--)
for (k = i - 1; k; k--)
if (max(max(sg, v1), 2 * sg + 1) <= i &&
max(max(sg, v2), 2 * sg + 1) <= j && max(sg, v3) <= k &&
k <= min(2 * sg, 2 * v3)) {
cout << i << endl << j << endl << k << endl;
return 0;
}
cout << -1 << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int jud(int person, int family) {
for (int i = 2 * family; i >= family; i--) {
if (person <= i && 2 * person >= i) return i;
}
return 0;
}
int get_in(int person, int family) {
for (int i = 2 * family; i >= family; i--) {
if (person <= i && 2 * person < i) {
return i;
}
}
return 0;
}
int ans[10];
int main() {
int v[10];
for (int i = 1; i <= 4; i++) {
cin >> v[i];
}
ans[1] = get_in(v[4], v[1]);
ans[2] = get_in(v[4], v[2]);
ans[3] = jud(v[4], v[3]);
if (ans[1] && ans[2] && ans[3] && ans[1] > ans[2] && ans[2] > ans[3]) {
cout << ans[1] << endl << ans[2] << endl << ans[3] << endl;
} else {
cout << -1 << endl;
}
}
| ### 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int jud(int person, int family) {
for (int i = 2 * family; i >= family; i--) {
if (person <= i && 2 * person >= i) return i;
}
return 0;
}
int get_in(int person, int family) {
for (int i = 2 * family; i >= family; i--) {
if (person <= i && 2 * person < i) {
return i;
}
}
return 0;
}
int ans[10];
int main() {
int v[10];
for (int i = 1; i <= 4; i++) {
cin >> v[i];
}
ans[1] = get_in(v[4], v[1]);
ans[2] = get_in(v[4], v[2]);
ans[3] = jud(v[4], v[3]);
if (ans[1] && ans[2] && ans[3] && ans[1] > ans[2] && ans[2] > ans[3]) {
cout << ans[1] << endl << ans[2] << endl << ans[3] << endl;
} else {
cout << -1 << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const long long mod = 10007;
const int N = 3000;
int main() {
int v[10];
int ans[10];
for (int i = 1; i <= 4; i++) {
scanf("%d", &v[i]);
ans[i] = v[i];
}
if (v[3] * 2 < v[4] || v[1] <= v[4] || v[2] <= v[4] || v[3] > 2 * v[4]) {
printf("-1");
} else {
ans[3] = max(v[4], v[3]);
ans[2] = max(v[4] * 2 + 1, v[2]);
ans[1] = max(ans[2] + 1, v[1]);
for (int i = 1; i <= 3; i++) {
printf("%d\n", ans[i]);
}
}
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const long long mod = 10007;
const int N = 3000;
int main() {
int v[10];
int ans[10];
for (int i = 1; i <= 4; i++) {
scanf("%d", &v[i]);
ans[i] = v[i];
}
if (v[3] * 2 < v[4] || v[1] <= v[4] || v[2] <= v[4] || v[3] > 2 * v[4]) {
printf("-1");
} else {
ans[3] = max(v[4], v[3]);
ans[2] = max(v[4] * 2 + 1, v[2]);
ans[1] = max(ans[2] + 1, v[1]);
for (int i = 1; i <= 3; i++) {
printf("%d\n", ans[i]);
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
while (scanf("%d%d%d%d", &a, &b, &c, &d) == 4) {
int L1 = c, R1 = 2 * c, L2 = d, R2 = 2 * d;
if (L1 <= R2 && L2 <= R1 && R2 < 2 * b) {
int t1 = max(L1, L2);
printf("%d %d %d\n", 2 * a, 2 * b, t1);
} else
printf("-1\n");
}
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
while (scanf("%d%d%d%d", &a, &b, &c, &d) == 4) {
int L1 = c, R1 = 2 * c, L2 = d, R2 = 2 * d;
if (L1 <= R2 && L2 <= R1 && R2 < 2 * b) {
int t1 = max(L1, L2);
printf("%d %d %d\n", 2 * a, 2 * b, t1);
} else
printf("-1\n");
}
return 0;
}
``` |
#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, m;
int flag = 0;
int c1 = 0, c2 = 0, c3 = 0;
int t;
cin >> a >> b >> c >> m;
t = 0;
for (int i = min(c, m); i <= 2 * c && i <= 2 * m; i++) {
if ((m <= i && 2 * m >= i) && (c <= i && 2 * c >= i)) {
t = 1;
c3 = i;
break;
}
}
if (t == 0) {
flag = 1;
}
if (flag == 0) {
t = 0;
for (int i = b; i <= 2 * b; i++) {
if ((m <= i) && (b <= i && 2 * b >= i) && i > c3 && (2 * m < i)) {
t = 1;
c2 = i;
break;
}
}
if (t == 0) {
flag = 1;
}
}
if (flag == 0) {
t = 0;
for (int i = a; i <= 2 * a; i++) {
if ((m <= i) && (a <= i && 2 * a >= i) && i > c2 && (2 * m < i)) {
t = 1;
c1 = i;
break;
}
}
if (t == 0) {
flag = 1;
}
}
if (flag == 1) {
cout << "-1\n";
} else {
cout << c1 << '\n' << c2 << '\n' << c3 << '\n';
}
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#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, m;
int flag = 0;
int c1 = 0, c2 = 0, c3 = 0;
int t;
cin >> a >> b >> c >> m;
t = 0;
for (int i = min(c, m); i <= 2 * c && i <= 2 * m; i++) {
if ((m <= i && 2 * m >= i) && (c <= i && 2 * c >= i)) {
t = 1;
c3 = i;
break;
}
}
if (t == 0) {
flag = 1;
}
if (flag == 0) {
t = 0;
for (int i = b; i <= 2 * b; i++) {
if ((m <= i) && (b <= i && 2 * b >= i) && i > c3 && (2 * m < i)) {
t = 1;
c2 = i;
break;
}
}
if (t == 0) {
flag = 1;
}
}
if (flag == 0) {
t = 0;
for (int i = a; i <= 2 * a; i++) {
if ((m <= i) && (a <= i && 2 * a >= i) && i > c2 && (2 * m < i)) {
t = 1;
c1 = i;
break;
}
}
if (t == 0) {
flag = 1;
}
}
if (flag == 1) {
cout << "-1\n";
} else {
cout << c1 << '\n' << c2 << '\n' << c3 << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
for (int i = a; i <= 2 * a; ++i)
for (int j = b; j <= 2 * b && j < i; ++j)
for (int k = c; k <= 2 * c && k < j; ++k)
if (d <= i && d <= j && d <= k && d * 2 >= k && d * 2 < j &&
d * 2 < i) {
printf("%d\n%d\n%d", i, j, k);
return 0;
}
puts("-1");
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
for (int i = a; i <= 2 * a; ++i)
for (int j = b; j <= 2 * b && j < i; ++j)
for (int k = c; k <= 2 * c && k < j; ++k)
if (d <= i && d <= j && d <= k && d * 2 >= k && d * 2 < j &&
d * 2 < i) {
printf("%d\n%d\n%d", i, j, k);
return 0;
}
puts("-1");
}
``` |
#include <bits/stdc++.h>
using namespace std;
void read(bool out = 0) {}
bool can(long long man, long long car) { return man <= car; }
bool like(long long man, long long car) { return 2 * man >= car; }
int main() {
long long a, b, c, k;
cin >> a >> b >> c >> k;
for (int i = 1; i < 201; i++)
for (int j = 1; j < 201; j++)
for (int l = 1; l < 201; l++) {
if (i > j && j > l) {
if (can(a, i) && like(a, i)) {
if (can(b, j) && like(b, j)) {
if (can(c, l) && like(c, l)) {
if (can(k, i) && can(k, j) && can(k, l)) {
if (like(k, l) && !like(k, i) && !like(k, j)) {
return cout << i << "\n" << j << "\n" << l << "\n", 0;
}
}
}
}
}
}
}
cout << -1;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void read(bool out = 0) {}
bool can(long long man, long long car) { return man <= car; }
bool like(long long man, long long car) { return 2 * man >= car; }
int main() {
long long a, b, c, k;
cin >> a >> b >> c >> k;
for (int i = 1; i < 201; i++)
for (int j = 1; j < 201; j++)
for (int l = 1; l < 201; l++) {
if (i > j && j > l) {
if (can(a, i) && like(a, i)) {
if (can(b, j) && like(b, j)) {
if (can(c, l) && like(c, l)) {
if (can(k, i) && can(k, j) && can(k, l)) {
if (like(k, l) && !like(k, i) && !like(k, j)) {
return cout << i << "\n" << j << "\n" << l << "\n", 0;
}
}
}
}
}
}
}
cout << -1;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int x, y;
int ap[100005], b[20000001] = {0};
int main() {
int b1, b2, b3, m;
cin >> b1 >> b2 >> b3 >> m;
for (int i = 0; i < 250; i++) {
for (int j = 0; j < 250; j++) {
for (int h = 0; h < 250; h++) {
if (i >= max(b1, m) && j >= max(b2, m) && h >= max(b3, m)) {
if (i <= 2 * b1 && j <= 2 * b2 && h <= 2 * b3) {
if (i > 2 * m && j > 2 * m) {
if (h <= 2 * m) {
if (i > j && j > h) {
cout << i << endl;
cout << j << endl;
cout << h << endl;
return 0;
}
}
}
}
}
}
}
}
cout << -1 << endl;
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int x, y;
int ap[100005], b[20000001] = {0};
int main() {
int b1, b2, b3, m;
cin >> b1 >> b2 >> b3 >> m;
for (int i = 0; i < 250; i++) {
for (int j = 0; j < 250; j++) {
for (int h = 0; h < 250; h++) {
if (i >= max(b1, m) && j >= max(b2, m) && h >= max(b3, m)) {
if (i <= 2 * b1 && j <= 2 * b2 && h <= 2 * b3) {
if (i > 2 * m && j > 2 * m) {
if (h <= 2 * m) {
if (i > j && j > h) {
cout << i << endl;
cout << j << endl;
cout << h << endl;
return 0;
}
}
}
}
}
}
}
}
cout << -1 << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int INF = 1000000007;
const int N = 100000 + 7;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j < i; j++) {
for (int k = 1; k < i; k++) {
if (a <= i && i <= 2 * a && b <= j && j <= 2 * b && c <= k &&
k <= 2 * c && d <= k && k <= 2 * d && j > 2 * d) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return 0;
}
}
}
}
cout << -1 << endl;
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int INF = 1000000007;
const int N = 100000 + 7;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j < i; j++) {
for (int k = 1; k < i; k++) {
if (a <= i && i <= 2 * a && b <= j && j <= 2 * b && c <= k &&
k <= 2 * c && d <= k && k <= 2 * d && j > 2 * d) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
return 0;
}
}
}
}
cout << -1 << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
while (scanf("%d%d%d%d", &a, &b, &c, &d) == 4) {
for (int i = a; i <= 2 * a; i++) {
for (int j = b; j <= 2 * b && j < i; j++) {
for (int k = c; k <= 2 * c && k < j; k++) {
if (a <= i && b <= j && c <= k && 2 * a >= i && 2 * b >= j &&
2 * c >= k && d <= j && d <= i && d <= k && 2 * d >= k &&
2 * d < i && 2 * d < j) {
cout << i << ' ' << j << ' ' << k << endl;
goto OK;
}
}
}
}
cout << -1 << endl;
continue;
OK:
continue;
}
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
while (scanf("%d%d%d%d", &a, &b, &c, &d) == 4) {
for (int i = a; i <= 2 * a; i++) {
for (int j = b; j <= 2 * b && j < i; j++) {
for (int k = c; k <= 2 * c && k < j; k++) {
if (a <= i && b <= j && c <= k && 2 * a >= i && 2 * b >= j &&
2 * c >= k && d <= j && d <= i && d <= k && 2 * d >= k &&
2 * d < i && 2 * d < j) {
cout << i << ' ' << j << ' ' << k << endl;
goto OK;
}
}
}
}
cout << -1 << endl;
continue;
OK:
continue;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
scanf("%d%d%d%d", &v1, &v2, &v3, &v4);
for (int a1 = 0; a1 <= 200; a1++)
for (int b1 = 0; b1 < a1; b1++)
for (int c1 = 0; c1 < b1; c1++)
if (a1 >= v1 && b1 >= v2 && c1 >= v3 && a1 <= v1 * 2 && b1 <= v2 * 2 &&
c1 <= v3 * 2 && v4 <= a1 && v4 <= b1 && v4 <= c1 && v4 * 2 < a1 &&
v4 * 2 < b1 && v4 * 2 >= c1) {
printf("%d\n%d\n%d\n", a1, b1, c1);
return 0;
}
printf("-1\n");
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
scanf("%d%d%d%d", &v1, &v2, &v3, &v4);
for (int a1 = 0; a1 <= 200; a1++)
for (int b1 = 0; b1 < a1; b1++)
for (int c1 = 0; c1 < b1; c1++)
if (a1 >= v1 && b1 >= v2 && c1 >= v3 && a1 <= v1 * 2 && b1 <= v2 * 2 &&
c1 <= v3 * 2 && v4 <= a1 && v4 <= b1 && v4 <= c1 && v4 * 2 < a1 &&
v4 * 2 < b1 && v4 * 2 >= c1) {
printf("%d\n%d\n%d\n", a1, b1, c1);
return 0;
}
printf("-1\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie();
cout.tie();
;
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
for (int i = 300; i >= 1; i--) {
if (v1 <= i && 2 * v1 >= i) {
for (int j = i - 1; j >= 1; j--) {
if (v2 <= j && 2 * v2 >= j) {
for (int q = j - 1; q >= 1; q--) {
if (v3 <= q && 2 * v3 >= q) {
if (v4 <= q && 2 * v4 >= q && 2 * v4 < j && 2 * v4 < i) {
return cout << i << endl << j << endl << q, 0;
}
}
}
}
}
}
}
cout << -1;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie();
cout.tie();
;
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
for (int i = 300; i >= 1; i--) {
if (v1 <= i && 2 * v1 >= i) {
for (int j = i - 1; j >= 1; j--) {
if (v2 <= j && 2 * v2 >= j) {
for (int q = j - 1; q >= 1; q--) {
if (v3 <= q && 2 * v3 >= q) {
if (v4 <= q && 2 * v4 >= q && 2 * v4 < j && 2 * v4 < i) {
return cout << i << endl << j << endl << q, 0;
}
}
}
}
}
}
}
cout << -1;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long sum(long long a) { return a * (a + 1) / 2; }
int main() {
int a, b, c, m;
cin >> a >> b >> c >> m;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j <= 200; j++) {
for (int k = 1; k <= 200; k++) {
if (i >= a && i <= 2 * a && j >= b && j <= 2 * b && k >= c &&
k <= 2 * c && k >= m && k <= 2 * m && i > j && j > k && i > 2 * m &&
j > 2 * m) {
cout << i << endl << j << endl << k << endl;
return 0;
}
}
}
}
cout << -1 << endl;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long sum(long long a) { return a * (a + 1) / 2; }
int main() {
int a, b, c, m;
cin >> a >> b >> c >> m;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j <= 200; j++) {
for (int k = 1; k <= 200; k++) {
if (i >= a && i <= 2 * a && j >= b && j <= 2 * b && k >= c &&
k <= 2 * c && k >= m && k <= 2 * m && i > j && j > k && i > 2 * m &&
j > 2 * m) {
cout << i << endl << j << endl << k << endl;
return 0;
}
}
}
}
cout << -1 << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long pwr(long long base, long long p, long long mod = 1000000000) {
long long ans = 1;
while (p) {
if (p % 2 == 1) ans = (ans * base) % 1000000000;
base = (base * base) % 1000000000;
p /= 2;
}
return ans;
}
long long gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
return (a.first > b.first);
}
const long long N = 5e5 + 5;
int main() {
int p, m, s, ma;
cin >> p >> m >> s >> ma;
if (s > 2 * ma || 2 * s < ma || m <= ma) {
cout << -1 << endl;
} else {
cout << 2 * p << endl << 2 * m << endl << max(s, ma) << endl;
}
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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long pwr(long long base, long long p, long long mod = 1000000000) {
long long ans = 1;
while (p) {
if (p % 2 == 1) ans = (ans * base) % 1000000000;
base = (base * base) % 1000000000;
p /= 2;
}
return ans;
}
long long gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
return (a.first > b.first);
}
const long long N = 5e5 + 5;
int main() {
int p, m, s, ma;
cin >> p >> m >> s >> ma;
if (s > 2 * ma || 2 * s < ma || m <= ma) {
cout << -1 << endl;
} else {
cout << 2 * p << endl << 2 * m << endl << max(s, ma) << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
bool check(int i, int j, int k) {
if (i >= max(v1, vm) && j >= max(v2, vm) && k >= max(v3, vm)) {
if (i <= 2 * v1 && j <= 2 * v2 && k <= 2 * v3) {
if (i > 2 * vm && j > 2 * vm) {
if (k <= 2 * vm) {
if (i > j && j > k) {
return true;
} else {
return false;
}
} else {
return false;
}
} else {
return false;
}
} else {
return false;
}
} else {
return false;
}
}
int main() {
cin >> v1 >> v2 >> v3 >> vm;
bool ans = false;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j <= 200; j++) {
for (int k = 0; k <= 200; k++) {
if (check(i, j, k)) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
ans = true;
return 0;
}
}
}
}
if (!ans) {
cout << -1 << endl;
}
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
bool check(int i, int j, int k) {
if (i >= max(v1, vm) && j >= max(v2, vm) && k >= max(v3, vm)) {
if (i <= 2 * v1 && j <= 2 * v2 && k <= 2 * v3) {
if (i > 2 * vm && j > 2 * vm) {
if (k <= 2 * vm) {
if (i > j && j > k) {
return true;
} else {
return false;
}
} else {
return false;
}
} else {
return false;
}
} else {
return false;
}
} else {
return false;
}
}
int main() {
cin >> v1 >> v2 >> v3 >> vm;
bool ans = false;
for (int i = 1; i <= 200; i++) {
for (int j = 1; j <= 200; j++) {
for (int k = 0; k <= 200; k++) {
if (check(i, j, k)) {
cout << i << endl;
cout << j << endl;
cout << k << endl;
ans = true;
return 0;
}
}
}
}
if (!ans) {
cout << -1 << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
const int INF = INT_MAX;
const long long INFLL = (long long)1e18;
const double PI = acos(-1.0);
template <class T>
T gcd(T a, T b) {
return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
return a / gcd(a, b) * b;
}
const int N = (int)1e5 + 5;
const int P = 1 << 18;
int a, b, c;
int m;
int main() {
ios::sync_with_stdio(false);
cout.setf(ios::fixed | ios::showpoint);
cout.precision(8);
cin >> a >> b >> c >> m;
int A = a * 2, B = b * 2, C = c;
if (m <= B && m * 2 >= B) return !printf("-1\n");
if (m > C * 2) return !printf("-1\n");
if (C % 2 == 1)
C = C / 2 + 1;
else
C /= 2;
if (m * 2 < C) return !printf("-1\n");
C = -1;
for (int i = c; i <= c * 2; i++) {
if (m <= i && 2 * m >= i) {
C = i;
break;
}
}
if (C == -1) return !printf("-1\n");
cout << A << '\n' << B << '\n' << C << 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
const int INF = INT_MAX;
const long long INFLL = (long long)1e18;
const double PI = acos(-1.0);
template <class T>
T gcd(T a, T b) {
return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
return a / gcd(a, b) * b;
}
const int N = (int)1e5 + 5;
const int P = 1 << 18;
int a, b, c;
int m;
int main() {
ios::sync_with_stdio(false);
cout.setf(ios::fixed | ios::showpoint);
cout.precision(8);
cin >> a >> b >> c >> m;
int A = a * 2, B = b * 2, C = c;
if (m <= B && m * 2 >= B) return !printf("-1\n");
if (m > C * 2) return !printf("-1\n");
if (C % 2 == 1)
C = C / 2 + 1;
else
C /= 2;
if (m * 2 < C) return !printf("-1\n");
C = -1;
for (int i = c; i <= c * 2; i++) {
if (m <= i && 2 * m >= i) {
C = i;
break;
}
}
if (C == -1) return !printf("-1\n");
cout << A << '\n' << B << '\n' << C << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
long long power(long long a, long long b) {
long long ret = 1;
while (b) {
if (b & 1) ret *= a;
a *= a;
if (ret >= MOD) ret %= MOD;
if (a >= MOD) a %= MOD;
b >>= 1;
}
return ret;
}
long long inv(long long x) { return power(x, MOD - 2); }
const int N = 1e5 + 5;
int v1, v2, v3, vm;
int main() {
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> v1 >> v2 >> v3 >> vm;
for (int b1 = 0; b1 <= 200; b1++) {
for (int b2 = 0; b2 <= 200; b2++) {
for (int b3 = 0; b3 <= 200; b3++) {
if (b1 > b2 and b2 > b3 and v1 <= b1 and 2 * v1 >= b1 and v2 <= b2 and
2 * v2 >= b2 and v3 <= b3 and 2 * v3 >= b3 and
(b3 >= vm and 2 * vm >= b3 and 2 * vm < b2)) {
cout << b1 << "\n" << b2 << "\n" << b3;
return 0;
}
}
}
}
cout << "-1";
}
| ### 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 car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
long long power(long long a, long long b) {
long long ret = 1;
while (b) {
if (b & 1) ret *= a;
a *= a;
if (ret >= MOD) ret %= MOD;
if (a >= MOD) a %= MOD;
b >>= 1;
}
return ret;
}
long long inv(long long x) { return power(x, MOD - 2); }
const int N = 1e5 + 5;
int v1, v2, v3, vm;
int main() {
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> v1 >> v2 >> v3 >> vm;
for (int b1 = 0; b1 <= 200; b1++) {
for (int b2 = 0; b2 <= 200; b2++) {
for (int b3 = 0; b3 <= 200; b3++) {
if (b1 > b2 and b2 > b3 and v1 <= b1 and 2 * v1 >= b1 and v2 <= b2 and
2 * v2 >= b2 and v3 <= b3 and 2 * v3 >= b3 and
(b3 >= vm and 2 * vm >= b3 and 2 * vm < b2)) {
cout << b1 << "\n" << b2 << "\n" << b3;
return 0;
}
}
}
}
cout << "-1";
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int n = 3;
int v[n], vm, s[n];
bool dfs(int p) {
if (p == 3) return true;
for (int i = v[p]; i <= 2 * v[p]; ++i) {
s[p] = i;
if (p) {
if (2 * vm < i && i > s[p - 1]) {
if (dfs(p + 1)) return true;
}
} else {
if (vm <= i && 2 * vm >= i) {
if (dfs(p + 1)) return true;
}
}
}
return false;
}
void solve() {
if (dfs(0)) {
printf("%d\n%d\n%d\n", s[2], s[1], s[0]);
} else {
printf("-1\n");
}
}
int main() {
scanf("%d%d%d%d", &v[2], &v[1], &v[0], &vm);
solve();
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int n = 3;
int v[n], vm, s[n];
bool dfs(int p) {
if (p == 3) return true;
for (int i = v[p]; i <= 2 * v[p]; ++i) {
s[p] = i;
if (p) {
if (2 * vm < i && i > s[p - 1]) {
if (dfs(p + 1)) return true;
}
} else {
if (vm <= i && 2 * vm >= i) {
if (dfs(p + 1)) return true;
}
}
}
return false;
}
void solve() {
if (dfs(0)) {
printf("%d\n%d\n%d\n", s[2], s[1], s[0]);
} else {
printf("-1\n");
}
}
int main() {
scanf("%d%d%d%d", &v[2], &v[1], &v[0], &vm);
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class T>
void rd(T& ret) {
ret = 0;
bool ok = 0, u = 0;
for (;;) {
int c = getchar();
if (c >= '0' && c <= '9')
ret = (ret << 3) + (ret << 1) + c - '0', ok = 1;
else if (c == '-')
u = 1;
else if (ok) {
if (u) ret *= -1;
return;
}
}
}
long long pow_mod(long long p, long long n, long long mod) {
long long ret = 1;
for (; n; n >>= 1) {
if (n & 1) ret = ret * p % mod;
p = p * p % mod;
}
return ret;
}
template <class T>
bool chmin(T& a, const T& b) {
return b < a ? a = b, 1 : 0;
}
template <class T>
bool chmax(T& a, const T& b) {
return b > a ? a = b, 1 : 0;
}
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int a, b, c, d;
bool climb(int a, int i) { return i >= a; }
bool like(int a, int i) { return (climb(a, i) && (2 * a >= i)); }
bool ok(int i, int j, int k) {
if (!(climb(a, i) && like(a, i))) return 0;
if (!(climb(b, j) && like(b, j))) return 0;
if (!(climb(c, k) && like(c, k))) return 0;
if (!(climb(d, k) && like(d, k))) return 0;
if (!climb(d, i)) return 0;
if (!climb(d, j)) return 0;
if (!climb(d, k)) return 0;
if (like(d, i)) return 0;
if (like(d, j)) return 0;
if (!like(d, k)) return 0;
return 1;
}
int main() {
cin >> a >> b >> c >> d;
for (int i = (1); i < (210); i++)
for (int j = (1); j < (i); j++)
for (int k = (1); k < (j); k++) {
if (ok(i, j, k)) {
cout << i << " " << j << " " << k << endl;
return 0;
}
}
puts("-1");
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
void rd(T& ret) {
ret = 0;
bool ok = 0, u = 0;
for (;;) {
int c = getchar();
if (c >= '0' && c <= '9')
ret = (ret << 3) + (ret << 1) + c - '0', ok = 1;
else if (c == '-')
u = 1;
else if (ok) {
if (u) ret *= -1;
return;
}
}
}
long long pow_mod(long long p, long long n, long long mod) {
long long ret = 1;
for (; n; n >>= 1) {
if (n & 1) ret = ret * p % mod;
p = p * p % mod;
}
return ret;
}
template <class T>
bool chmin(T& a, const T& b) {
return b < a ? a = b, 1 : 0;
}
template <class T>
bool chmax(T& a, const T& b) {
return b > a ? a = b, 1 : 0;
}
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int a, b, c, d;
bool climb(int a, int i) { return i >= a; }
bool like(int a, int i) { return (climb(a, i) && (2 * a >= i)); }
bool ok(int i, int j, int k) {
if (!(climb(a, i) && like(a, i))) return 0;
if (!(climb(b, j) && like(b, j))) return 0;
if (!(climb(c, k) && like(c, k))) return 0;
if (!(climb(d, k) && like(d, k))) return 0;
if (!climb(d, i)) return 0;
if (!climb(d, j)) return 0;
if (!climb(d, k)) return 0;
if (like(d, i)) return 0;
if (like(d, j)) return 0;
if (!like(d, k)) return 0;
return 1;
}
int main() {
cin >> a >> b >> c >> d;
for (int i = (1); i < (210); i++)
for (int j = (1); j < (i); j++)
for (int k = (1); k < (j); k++) {
if (ok(i, j, k)) {
cout << i << " " << j << " " << k << endl;
return 0;
}
}
puts("-1");
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:5000000000")
const unsigned long long mod = 1000000007;
long long Inf = (long long)2e9;
long long LINF = (long long)1e18 + 1e17;
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int i = a; i <= 2 * a; i++) {
for (int j = b; j <= 2 * b; j++) {
if (j < i) {
int x_max = min(j - 1, 2 * c), x_min = c;
if (2 * d < x_min || d > x_max) continue;
if (2 * d >= j) continue;
for (int k = x_min; k <= x_max; k++) {
if (2 * d >= k && k >= d && 2 * k < j) {
cout << i << " " << j << " " << k;
return 0;
}
}
}
}
}
puts("-1");
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:5000000000")
const unsigned long long mod = 1000000007;
long long Inf = (long long)2e9;
long long LINF = (long long)1e18 + 1e17;
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
for (int i = a; i <= 2 * a; i++) {
for (int j = b; j <= 2 * b; j++) {
if (j < i) {
int x_max = min(j - 1, 2 * c), x_min = c;
if (2 * d < x_min || d > x_max) continue;
if (2 * d >= j) continue;
for (int k = x_min; k <= x_max; k++) {
if (2 * d >= k && k >= d && 2 * k < j) {
cout << i << " " << j << " " << k;
return 0;
}
}
}
}
}
puts("-1");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int const N = 5005;
int mod = 1000000007;
int main() {
int v1, v2, v3, vm;
cin >> v3 >> v2 >> v1 >> vm;
if ((v1 > 2 * vm or vm > 2 * v1) or 2 * v3 <= 2 * vm or 2 * v2 <= 2 * vm)
cout << -1 << endl;
else {
cout << v3 * 2 << endl;
cout << v2 * 2 << endl;
cout << max(vm, v1) << endl;
}
return 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int const N = 5005;
int mod = 1000000007;
int main() {
int v1, v2, v3, vm;
cin >> v3 >> v2 >> v1 >> vm;
if ((v1 > 2 * vm or vm > 2 * v1) or 2 * v3 <= 2 * vm or 2 * v2 <= 2 * vm)
cout << -1 << endl;
else {
cout << v3 * 2 << endl;
cout << v2 * 2 << endl;
cout << max(vm, v1) << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
int f, m, s, ma;
int main() {
scanf("%d %d %d %d", &f, &m, &s, &ma);
int fm, fmi, mm, sm, smi, mam, mmi;
fm = 2 * f;
mm = 2 * m;
sm = 2 * s;
mam = m * 2;
if (ma > fm || ma > mm || ma > sm || 2 * s < ma || 2 * ma < s ||
(2 * m <= 2 * ma)) {
printf("-1");
return 0;
}
int fc, mc, sc;
fc = 2 * f;
mc = 2 * m;
if (ma <= s)
sc = s;
else
sc = ma;
if (sc >= mc) {
printf("-1");
return 0;
}
printf("%d\n%d\n%d", fc, mc, sc);
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
int f, m, s, ma;
int main() {
scanf("%d %d %d %d", &f, &m, &s, &ma);
int fm, fmi, mm, sm, smi, mam, mmi;
fm = 2 * f;
mm = 2 * m;
sm = 2 * s;
mam = m * 2;
if (ma > fm || ma > mm || ma > sm || 2 * s < ma || 2 * ma < s ||
(2 * m <= 2 * ma)) {
printf("-1");
return 0;
}
int fc, mc, sc;
fc = 2 * f;
mc = 2 * m;
if (ma <= s)
sc = s;
else
sc = ma;
if (sc >= mc) {
printf("-1");
return 0;
}
printf("%d\n%d\n%d", fc, mc, sc);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
if (2 * v3 < v4 || 2 * v4 < v3 || v4 >= v2) {
cout << "-1";
} else {
cout << 2 * v1 << endl << 2 * v2 << endl << max(v3, v4);
}
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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v4;
cin >> v1 >> v2 >> v3 >> v4;
if (2 * v3 < v4 || 2 * v4 < v3 || v4 >= v2) {
cout << "-1";
} else {
cout << 2 * v1 << endl << 2 * v2 << endl << max(v3, v4);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 1;
vector<int> v[N];
int vis[N];
long long int cnt1 = 0, cnt2 = 0, cnt = 0, m, n, x, y, z, i, j, k, mini, d, a,
b, c, o;
int color[N] = {0};
long long int brr[N];
void dfs(int s, int t) {
vis[s] = 1;
color[s] = t;
for (int i = 0; i < v[s].size(); i++) {
if (!vis[v[s][i]]) dfs(v[s][i], t);
}
}
int main() {
cin >> a >> b >> c >> d;
for (int i = a; i <= 2 * a; i++) {
for (int j = b; j <= 2 * b; j++) {
for (int k = c; k <= 2 * c; k++) {
if (i > j && j > k && k >= d && 2 * d >= k && 2 * d < j) {
cout << i << endl << j << endl << k << endl;
return 0;
}
}
}
}
cout << -1;
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 1;
vector<int> v[N];
int vis[N];
long long int cnt1 = 0, cnt2 = 0, cnt = 0, m, n, x, y, z, i, j, k, mini, d, a,
b, c, o;
int color[N] = {0};
long long int brr[N];
void dfs(int s, int t) {
vis[s] = 1;
color[s] = t;
for (int i = 0; i < v[s].size(); i++) {
if (!vis[v[s][i]]) dfs(v[s][i], t);
}
}
int main() {
cin >> a >> b >> c >> d;
for (int i = a; i <= 2 * a; i++) {
for (int j = b; j <= 2 * b; j++) {
for (int k = c; k <= 2 * c; k++) {
if (i > j && j > k && k >= d && 2 * d >= k && 2 * d < j) {
cout << i << endl << j << endl << k << endl;
return 0;
}
}
}
}
cout << -1;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
int f, m, s, mas;
cin >> f >> m >> s >> mas;
if (mas > 2 * s || s > 2 * mas) {
cout << -1;
return 0;
}
int sc = s;
while (sc < mas) ++sc;
if (sc >= m) {
cout << -1;
return 0;
}
cout << 2 * f << endl << 2 * m << endl;
cout << sc << endl;
return 0;
}
| ### 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 and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
int f, m, s, mas;
cin >> f >> m >> s >> mas;
if (mas > 2 * s || s > 2 * mas) {
cout << -1;
return 0;
}
int sc = s;
while (sc < mas) ++sc;
if (sc >= m) {
cout << -1;
return 0;
}
cout << 2 * f << endl << 2 * m << endl;
cout << sc << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int Father, Mother, Son, Masha, sizeMasha;
int ans1, ans2, ans3;
int main() {
cin >> Father >> Mother >> Son >> Masha;
for (int sizeFather = Father; sizeFather <= 2 * Father; sizeFather++) {
for (int sizeMother = Mother; sizeMother <= 2 * Mother; sizeMother++) {
for (int sizeSon = Son; sizeSon <= 2 * Son; sizeSon++) {
sizeMasha = 2 * Masha;
if (sizeFather > sizeMother && sizeMother > sizeSon) {
if (sizeMasha >= sizeSon && sizeMasha < sizeMother &&
sizeMasha < sizeFather) {
if (Masha <= sizeSon && Masha <= sizeMother &&
Masha <= sizeFather) {
ans1 = sizeFather, ans2 = sizeMother, ans3 = sizeSon;
cout << ans1 << "\n" << ans2 << "\n" << ans3 << "\n";
return 0;
}
}
}
}
}
}
cout << "-1\n";
return 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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int Father, Mother, Son, Masha, sizeMasha;
int ans1, ans2, ans3;
int main() {
cin >> Father >> Mother >> Son >> Masha;
for (int sizeFather = Father; sizeFather <= 2 * Father; sizeFather++) {
for (int sizeMother = Mother; sizeMother <= 2 * Mother; sizeMother++) {
for (int sizeSon = Son; sizeSon <= 2 * Son; sizeSon++) {
sizeMasha = 2 * Masha;
if (sizeFather > sizeMother && sizeMother > sizeSon) {
if (sizeMasha >= sizeSon && sizeMasha < sizeMother &&
sizeMasha < sizeFather) {
if (Masha <= sizeSon && Masha <= sizeMother &&
Masha <= sizeFather) {
ans1 = sizeFather, ans2 = sizeMother, ans3 = sizeSon;
cout << ans1 << "\n" << ans2 << "\n" << ans3 << "\n";
return 0;
}
}
}
}
}
}
cout << "-1\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
int main() {
cin >> v1 >> v2 >> v3 >> vm;
if (vm > v3 * 2 || vm >= v2 || vm * 2 < v3) {
cout << -1;
return 0;
} else {
cout << v1 * 2 << " " << v2 * 2 << " " << max(v3, vm);
}
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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int v1, v2, v3, vm;
int main() {
cin >> v1 >> v2 >> v3 >> vm;
if (vm > v3 * 2 || vm >= v2 || vm * 2 < v3) {
cout << -1;
return 0;
} else {
cout << v1 * 2 << " " << v2 * 2 << " " << max(v3, vm);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d;
int ansa, ansb, ansc;
int read() {
int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
int main() {
a = read();
b = read();
c = read();
d = read();
if (c * 2 < d || d * 2 < c || a <= d || b <= d) {
printf("-1");
return 0;
}
printf("%d\n%d\n%d", a * 2, b * 2, min(c * 2, d * 2));
return 0;
}
| ### 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 the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d;
int ansa, ansb, ansc;
int read() {
int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
int main() {
a = read();
b = read();
c = read();
d = read();
if (c * 2 < d || d * 2 < c || a <= d || b <= d) {
printf("-1");
return 0;
}
printf("%d\n%d\n%d", a * 2, b * 2, min(c * 2, d * 2));
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v;
cin >> v1 >> v2 >> v3 >> v;
int second = -1, m = -1, l = -1;
for (int i = v; i <= 230; i++) {
for (int j = i + 1; j <= 230; j++) {
for (int k = j + 1; k <= 230; k++) {
if (2 * v < j && 2 * v < k && 2 * v >= i && 2 * v3 >= i && i >= v3 &&
2 * v2 >= j && 2 * v1 >= k && j >= v2 && k >= v1) {
second = i;
m = j;
l = k;
break;
}
}
if (second != -1) break;
}
if (second != -1) break;
}
if (second == -1)
cout << -1;
else {
cout << l << "\n";
cout << m << "\n";
cout << second << "\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 smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Examples
Input
50 30 10 10
Output
50
30
10
Input
100 50 10 21
Output
-1
Note
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int v1, v2, v3, v;
cin >> v1 >> v2 >> v3 >> v;
int second = -1, m = -1, l = -1;
for (int i = v; i <= 230; i++) {
for (int j = i + 1; j <= 230; j++) {
for (int k = j + 1; k <= 230; k++) {
if (2 * v < j && 2 * v < k && 2 * v >= i && 2 * v3 >= i && i >= v3 &&
2 * v2 >= j && 2 * v1 >= k && j >= v2 && k >= v1) {
second = i;
m = j;
l = k;
break;
}
}
if (second != -1) break;
}
if (second != -1) break;
}
if (second == -1)
cout << -1;
else {
cout << l << "\n";
cout << m << "\n";
cout << second << "\n";
}
}
``` |
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