output
stringlengths 52
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stringlengths 296
182k
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|---|---|
#include <bits/stdc++.h>
using namespace std;
unsigned long long int fac(unsigned long long int n) {
return (n == 1 || n == 0) ? 1 : n * fac(n - 1);
}
unsigned long long int c(unsigned long long int n, unsigned long long int r) {
unsigned long long int k = fac(r);
return fac(n) / (k * k);
}
int main() {
unsigned long long int n;
cin >> n;
n--;
cout << c(2 * n, n);
return 0;
}
|
### Prompt
Please create a solution in Cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
unsigned long long int fac(unsigned long long int n) {
return (n == 1 || n == 0) ? 1 : n * fac(n - 1);
}
unsigned long long int c(unsigned long long int n, unsigned long long int r) {
unsigned long long int k = fac(r);
return fac(n) / (k * k);
}
int main() {
unsigned long long int n;
cin >> n;
n--;
cout << c(2 * n, n);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j, k = INT_MIN;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (a[i][j] > k) k = a[i][j];
}
}
cout << k;
return 0;
}
|
### Prompt
Please create a solution in Cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j, k = INT_MIN;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (a[i][j] > k) k = a[i][j];
}
}
cout << k;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n == 1) cout << "1";
if (n == 2) cout << "2";
if (n == 3) cout << "6";
if (n == 4) cout << "20";
if (n == 5) cout << "70";
if (n == 6) cout << "252";
if (n == 7) cout << "924";
if (n == 8) cout << "3432";
if (n == 9) cout << "12870";
if (n == 10) cout << "48620";
return 0;
}
|
### Prompt
Construct a cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n == 1) cout << "1";
if (n == 2) cout << "2";
if (n == 3) cout << "6";
if (n == 4) cout << "20";
if (n == 5) cout << "70";
if (n == 6) cout << "252";
if (n == 7) cout << "924";
if (n == 8) cout << "3432";
if (n == 9) cout << "12870";
if (n == 10) cout << "48620";
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
int i, j;
int *table;
scanf("%d", &n);
table = (int *)malloc(n * n * sizeof(int));
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
*(table + i * n + j) = 1;
else
*(table + i * n + j) =
*(table + i * n + j - 1) + *(table + (i - 1) * n + j);
}
}
printf("%d", *(table + (i - 1) * n + j - 1));
return 0;
}
|
### Prompt
Generate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n;
int i, j;
int *table;
scanf("%d", &n);
table = (int *)malloc(n * n * sizeof(int));
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
*(table + i * n + j) = 1;
else
*(table + i * n + j) =
*(table + i * n + j - 1) + *(table + (i - 1) * n + j);
}
}
printf("%d", *(table + (i - 1) * n + j - 1));
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
scanf("%d", &n);
int m = 2 * (n - 1);
long long ans = 1;
for (int i = 1; i <= m; i++) ans *= i;
for (int i = 1; i <= (n - 1); i++) ans /= i, ans /= i;
cout << ans << endl;
}
|
### Prompt
Develop a solution in Cpp to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
scanf("%d", &n);
int m = 2 * (n - 1);
long long ans = 1;
for (int i = 1; i <= m; i++) ans *= i;
for (int i = 1; i <= (n - 1); i++) ans /= i, ans /= i;
cout << ans << endl;
}
```
|
#include <bits/stdc++.h>
int max[25][25];
int main() {
int n, i;
scanf("%d", &n);
for (i = 1; i < 25; i++) max[1][i] = 1;
for (i = 2; i < 25; i++)
for (int j = 1; j < 25; j++) max[i][j] = max[i][j - 1] + max[i - 1][j];
printf("%d\n", max[n][n]);
return 0;
}
|
### Prompt
Your task is to create a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int max[25][25];
int main() {
int n, i;
scanf("%d", &n);
for (i = 1; i < 25; i++) max[1][i] = 1;
for (i = 2; i < 25; i++)
for (int j = 1; j < 25; j++) max[i][j] = max[i][j - 1] + max[i - 1][j];
printf("%d\n", max[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n;
int a[20][20];
int get(int i, int j) {
if (i == 1 || j == 1) return 1;
return a[i - 1][j] + a[i][j - 1];
}
int main() {
cin >> n;
memset(a, 0, sizeof(a));
for (int i = 0; i <= n; i++) a[1][i] = 1;
for (int i = 0; i <= n; i++) a[i][1] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
a[i][j] = get(i, j);
}
}
cout << a[n][n] << endl;
return 0;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
int a[20][20];
int get(int i, int j) {
if (i == 1 || j == 1) return 1;
return a[i - 1][j] + a[i][j - 1];
}
int main() {
cin >> n;
memset(a, 0, sizeof(a));
for (int i = 0; i <= n; i++) a[1][i] = 1;
for (int i = 0; i <= n; i++) a[i][1] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
a[i][j] = get(i, j);
}
}
cout << a[n][n] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x, max = 0;
cin >> x;
int a[x][x];
for (int i = 0; i < x; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int j = 1; j < x; j++) {
for (int k = 1; k < x; k++) {
a[j][k] = a[j - 1][k] + a[j][k - 1];
if (a[j][k] > max) {
max = a[j][k];
}
}
}
if (x == 1) {
cout << "1";
} else {
cout << max;
}
return 0;
}
|
### Prompt
Generate a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x, max = 0;
cin >> x;
int a[x][x];
for (int i = 0; i < x; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int j = 1; j < x; j++) {
for (int k = 1; k < x; k++) {
a[j][k] = a[j - 1][k] + a[j][k - 1];
if (a[j][k] > max) {
max = a[j][k];
}
}
}
if (x == 1) {
cout << "1";
} else {
cout << max;
}
return 0;
}
```
|
#include <bits/stdc++.h>
int maze[15][15] = {0};
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= 10; i++) {
maze[1][i] = 1;
maze[i][1] = 1;
}
for (int i = 2; i <= 10; i++) {
for (int j = 2; j <= 10; j++) {
maze[i][j] = maze[i - 1][j] + maze[i][j - 1];
}
}
printf("%d\n", maze[n][n]);
return 0;
}
|
### Prompt
Please formulate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int maze[15][15] = {0};
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= 10; i++) {
maze[1][i] = 1;
maze[i][1] = 1;
}
for (int i = 2; i <= 10; i++) {
for (int j = 2; j <= 10; j++) {
maze[i][j] = maze[i - 1][j] + maze[i][j - 1];
}
}
printf("%d\n", maze[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a(int i, int j) {
if (i == 1 || j == 1) return 1;
return a(i - 1, j) + a(i, j - 1);
}
int main() {
int n;
scanf("%d", &n);
printf("%d\n", a(n, n));
return 0;
}
|
### Prompt
Please provide a CPP coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a(int i, int j) {
if (i == 1 || j == 1) return 1;
return a(i - 1, j) + a(i, j - 1);
}
int main() {
int n;
scanf("%d", &n);
printf("%d\n", a(n, n));
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int s[200][200], n;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) s[i][1] = 1, s[1][i] = 1;
for (int i = 2; i <= n; i++)
for (int j = 2; j <= n; j++) s[i][j] = s[i - 1][j] + s[i][j - 1];
printf("%d", s[n][n]);
return 0;
}
|
### Prompt
Develop a solution in Cpp to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int s[200][200], n;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) s[i][1] = 1, s[1][i] = 1;
for (int i = 2; i <= n; i++)
for (int j = 2; j <= n; j++) s[i][j] = s[i - 1][j] + s[i][j - 1];
printf("%d", s[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv) {
int C[20][20];
C[0][0] = 1;
for (int i = (0); i < (20); ++i) {
for (int j = (0); j < (20); ++j) {
if (i == 0 || j == 0)
C[i][j] = 1;
else {
C[i][j] = C[i - 1][j] + C[i][j - 1];
}
}
}
int N;
cin >> N;
int res = C[0][0];
for (int i = (0); i < (N); ++i) {
for (int j = (0); j < (N); ++j) {
res = max(res, C[i][j]);
}
}
printf("%d\n", res);
return 0;
}
|
### Prompt
Generate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv) {
int C[20][20];
C[0][0] = 1;
for (int i = (0); i < (20); ++i) {
for (int j = (0); j < (20); ++j) {
if (i == 0 || j == 0)
C[i][j] = 1;
else {
C[i][j] = C[i - 1][j] + C[i][j - 1];
}
}
}
int N;
cin >> N;
int res = C[0][0];
for (int i = (0); i < (N); ++i) {
for (int j = (0); j < (N); ++j) {
res = max(res, C[i][j]);
}
}
printf("%d\n", res);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long i = 0, num, n = 1, d = 1;
cin >> num;
for (i = 0; i < num - 1; i++) {
n *= (num + i);
d *= (1 + i);
}
cout << n / d;
return 0;
}
|
### Prompt
Develop a solution in Cpp to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long i = 0, num, n = 1, d = 1;
cin >> num;
for (i = 0; i < num - 1; i++) {
n *= (num + i);
d *= (1 + i);
}
cout << n / d;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int n1, long long int n2) {
if (n1 % n2 == 0) return n2;
return gcd(n2, n1 % n2);
}
long long int powmod(long long int base, long long int exponent) {
long long int ans = 1;
while (exponent) {
if (exponent & 1) ans = (ans * base);
base = (base * base);
exponent /= 2;
}
return ans;
}
void Sieve(int n) {
int i, j;
bool primes[n];
for (i = 0; i < n; i++) {
primes[i] = true;
}
for (i = 2; i * i <= n; i++) {
if (primes[i] == true) {
for (j = i * i; j <= n; j += i) {
primes[j] = false;
}
}
}
}
long long int modexp(long long int a, long long int b, long long int m) {
if (b == 0) return 1;
long long int temp = modexp(a, b / 2, m);
if (b % 2 == 1) return (((temp % m) * (temp % m)) % m * (a % m)) % m;
return ((temp % m) * (temp % m)) % m;
}
int factor(int n) {
int i, c = 0;
for (i = 1; i * i <= n; i++) {
if (i * i == n)
c += 1;
else if (n % i == 0)
c += 2;
}
return c;
}
bool isLucky(int n) {
int m = n, x;
while (m) {
x = m % 10;
if (x != 4 && x != 7) return false;
m /= 10;
}
return true;
}
int main() {
int A[10][10], i, j, n;
for (i = 0; i < 10; i++) {
A[0][i] = 1;
A[i][0] = 1;
}
for (i = 1; i < 10; i++) {
for (j = 1; j < 10; j++) {
A[i][j] = A[i - 1][j] + A[i][j - 1];
}
}
cin >> n;
cout << A[n - 1][n - 1] << '\n';
}
|
### Prompt
Please create a solution in cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int n1, long long int n2) {
if (n1 % n2 == 0) return n2;
return gcd(n2, n1 % n2);
}
long long int powmod(long long int base, long long int exponent) {
long long int ans = 1;
while (exponent) {
if (exponent & 1) ans = (ans * base);
base = (base * base);
exponent /= 2;
}
return ans;
}
void Sieve(int n) {
int i, j;
bool primes[n];
for (i = 0; i < n; i++) {
primes[i] = true;
}
for (i = 2; i * i <= n; i++) {
if (primes[i] == true) {
for (j = i * i; j <= n; j += i) {
primes[j] = false;
}
}
}
}
long long int modexp(long long int a, long long int b, long long int m) {
if (b == 0) return 1;
long long int temp = modexp(a, b / 2, m);
if (b % 2 == 1) return (((temp % m) * (temp % m)) % m * (a % m)) % m;
return ((temp % m) * (temp % m)) % m;
}
int factor(int n) {
int i, c = 0;
for (i = 1; i * i <= n; i++) {
if (i * i == n)
c += 1;
else if (n % i == 0)
c += 2;
}
return c;
}
bool isLucky(int n) {
int m = n, x;
while (m) {
x = m % 10;
if (x != 4 && x != 7) return false;
m /= 10;
}
return true;
}
int main() {
int A[10][10], i, j, n;
for (i = 0; i < 10; i++) {
A[0][i] = 1;
A[i][0] = 1;
}
for (i = 1; i < 10; i++) {
for (j = 1; j < 10; j++) {
A[i][j] = A[i - 1][j] + A[i][j - 1];
}
}
cin >> n;
cout << A[n - 1][n - 1] << '\n';
}
```
|
#include <bits/stdc++.h>
using namespace std;
int xd[] = {0, 1, 0, -1};
int yd[] = {1, 0, -1, 0};
int xx[] = {0, 1, 0, -1, 1, -1, 1, -1};
int yy[] = {1, 0, -1, 0, 1, 1, -1, -1};
int kx[] = {-2, -1, -2, -1, 1, 2, 2, 1};
int ky[] = {1, 2, -1, -2, -2, -1, 1, 2};
bool islucky(int x) {
int four = 0, seven = 0;
while (x) {
if (x % 10 != 7 && x % 10 != 4) return false;
if (x % 10 == 7)
seven++;
else
four++;
x = x / 10;
}
return four == seven;
}
bool is_prime(long long x) {
if (x == 1) return false;
for (int i = 2; i * i <= x; i++)
if (x % i == 0) return false;
return true;
}
long long C(int k, int l) {
if (l == k) return 1;
if (l * 2 > k) l = k - l;
long long res = 1, temp = l;
for (int i = k - l + 1; i <= k; i++) {
res *= i / temp;
temp--;
}
return res;
}
long long sumOfDigit(int x) {
int res = 0;
while (x) {
res += x % 10;
x /= 10;
}
return res;
}
uint64_t bigPow(uint64_t base, uint64_t p) {
uint64_t temp = 1;
while (p--) {
temp *= base;
}
return temp;
}
vector<int> soso;
bool prime[10000001] = {0};
void SieveOfEratosthenes(int n) {
soso.push_back(2);
for (int p = 3; p * p <= n; p += 2) {
if (prime[p] == false) {
for (int i = p * 2; i <= n; i += p) prime[i] = true;
}
}
for (int i = 3; i <= n; i += 2)
if (!prime[i]) soso.push_back(i);
}
vector<vector<int> > AdjList;
bool visited[100005];
void dfs(int s) {
if (visited[s]) return;
visited[s] = 1;
for (int i = 0; i < AdjList[s].size(); i++) dfs(AdjList[s][i]);
return;
}
long long gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
int n;
cin >> n;
int a[11][11];
for (int i = 0; i < n; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1];
}
|
### Prompt
In Cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int xd[] = {0, 1, 0, -1};
int yd[] = {1, 0, -1, 0};
int xx[] = {0, 1, 0, -1, 1, -1, 1, -1};
int yy[] = {1, 0, -1, 0, 1, 1, -1, -1};
int kx[] = {-2, -1, -2, -1, 1, 2, 2, 1};
int ky[] = {1, 2, -1, -2, -2, -1, 1, 2};
bool islucky(int x) {
int four = 0, seven = 0;
while (x) {
if (x % 10 != 7 && x % 10 != 4) return false;
if (x % 10 == 7)
seven++;
else
four++;
x = x / 10;
}
return four == seven;
}
bool is_prime(long long x) {
if (x == 1) return false;
for (int i = 2; i * i <= x; i++)
if (x % i == 0) return false;
return true;
}
long long C(int k, int l) {
if (l == k) return 1;
if (l * 2 > k) l = k - l;
long long res = 1, temp = l;
for (int i = k - l + 1; i <= k; i++) {
res *= i / temp;
temp--;
}
return res;
}
long long sumOfDigit(int x) {
int res = 0;
while (x) {
res += x % 10;
x /= 10;
}
return res;
}
uint64_t bigPow(uint64_t base, uint64_t p) {
uint64_t temp = 1;
while (p--) {
temp *= base;
}
return temp;
}
vector<int> soso;
bool prime[10000001] = {0};
void SieveOfEratosthenes(int n) {
soso.push_back(2);
for (int p = 3; p * p <= n; p += 2) {
if (prime[p] == false) {
for (int i = p * 2; i <= n; i += p) prime[i] = true;
}
}
for (int i = 3; i <= n; i += 2)
if (!prime[i]) soso.push_back(i);
}
vector<vector<int> > AdjList;
bool visited[100005];
void dfs(int s) {
if (visited[s]) return;
visited[s] = 1;
for (int i = 0; i < AdjList[s].size(); i++) dfs(AdjList[s][i]);
return;
}
long long gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
int n;
cin >> n;
int a[11][11];
for (int i = 0; i < n; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
int main() {
int n, i, a[200][200], j;
while (scanf("%d", &n) != EOF) {
for (i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (i = 2; i <= n; i++)
for (j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
printf("%d\n", a[n][n]);
}
return 0;
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n, i, a[200][200], j;
while (scanf("%d", &n) != EOF) {
for (i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (i = 2; i <= n; i++)
for (j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
printf("%d\n", a[n][n]);
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
signed long long int d[102][102];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) d[1][i] = d[i][1] = 1;
for (int i = 2; i <= n; ++i) {
for (int j = 2; j <= n; ++j) {
d[i][j] = d[i - 1][j] + d[i][j - 1];
}
}
cout << d[n][n];
}
|
### Prompt
In Cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
signed long long int d[102][102];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) d[1][i] = d[i][1] = 1;
for (int i = 2; i <= n; ++i) {
for (int j = 2; j <= n; ++j) {
d[i][j] = d[i - 1][j] + d[i][j - 1];
}
}
cout << d[n][n];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n;
int a[1005][1005];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
cin >> n;
for (int i = 1; i <= n; i++) a[1][i] = a[i][1] = 1;
for (int i = 2; i <= n; i++)
for (int j = 2; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
cout << a[n][n];
return 0;
}
|
### Prompt
Please formulate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
int a[1005][1005];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
cin >> n;
for (int i = 1; i <= n; i++) a[1][i] = a[i][1] = 1;
for (int i = 2; i <= n; i++)
for (int j = 2; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
cout << a[n][n];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int** arr = new int*[n];
for (int i = 0; i < n; i++) arr[i] = new int[n];
for (int i = 0; i < n; i++) {
arr[0][i] = 1;
arr[i][0] = 1;
}
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
cout << arr[n - 1][n - 1];
return 0;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int** arr = new int*[n];
for (int i = 0; i < n; i++) arr[i] = new int[n];
for (int i = 0; i < n; i++) {
arr[0][i] = 1;
arr[i][0] = 1;
}
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
cout << arr[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
long long n;
cin >> n;
long long a[10][10];
for (auto(i) = (0); (i) < (n); (i)++) a[0][i] = 1;
for (auto(i) = (0); (i) < (n); (i)++) a[i][0] = 1;
for (auto(i) = (1); (i) < (n); (i)++)
for (auto(j) = (1); (j) < (n); (j)++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Your task is to create a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
long long n;
cin >> n;
long long a[10][10];
for (auto(i) = (0); (i) < (n); (i)++) a[0][i] = 1;
for (auto(i) = (0); (i) < (n); (i)++) a[i][0] = 1;
for (auto(i) = (1); (i) < (n); (i)++)
for (auto(j) = (1); (j) < (n); (j)++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
void solution() {
int num;
cin >> num;
int row[num];
for (int i = 0; i < num; i++) {
row[i] = 1;
}
for (int j = 0; j < num - 1; j++) {
for (int i = 1; i < num; i++) {
row[i] += row[i - 1];
}
}
cout << row[num - 1];
}
int main() {
solution();
return 0;
}
|
### Prompt
Generate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solution() {
int num;
cin >> num;
int row[num];
for (int i = 0; i < num; i++) {
row[i] = 1;
}
for (int j = 0; j < num - 1; j++) {
for (int i = 1; i < num; i++) {
row[i] += row[i - 1];
}
}
cout << row[num - 1];
}
int main() {
solution();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[10][10];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
}
}
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Generate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[10][10];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
}
}
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m = 0;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
m = max(m, a[i][j]);
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
m = max(m, a[i][j]);
}
}
}
cout << m;
return 0;
}
|
### Prompt
Generate a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m = 0;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
m = max(m, a[i][j]);
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
m = max(m, a[i][j]);
}
}
}
cout << m;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long int fact(int n) {
if (n == 0) {
return 1;
}
return n * fact(n - 1);
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
int n, k;
cin >> n;
k = 2 * n - 2;
cout << fact(k) / (fact(k / 2) * fact(k / 2));
return 0;
}
|
### Prompt
In CPP, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int fact(int n) {
if (n == 0) {
return 1;
}
return n * fact(n - 1);
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
int n, k;
cin >> n;
k = 2 * n - 2;
cout << fact(k) / (fact(k / 2) * fact(k / 2));
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, res = 0, a[11][11];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) a[i][1] = a[1][i] = 1;
for (int i = 2; i <= n; i++)
for (int j = 2; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) res = max(res, a[i][j]);
cout << res;
}
|
### Prompt
Create a solution in cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, res = 0, a[11][11];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) a[i][1] = a[1][i] = 1;
for (int i = 2; i <= n; i++)
for (int j = 2; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) res = max(res, a[i][j]);
cout << res;
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
int ara[n][n];
int i, j;
for (i = 0; i < n; i++) {
ara[0][i] = 1;
}
for (i = 1; i < n; i++) {
ara[i][0] = 1;
for (j = 1; j < n; j++) {
ara[i][j] = ara[i - 1][j] + ara[i][j - 1];
}
}
printf("%d", ara[n - 1][n - 1]);
}
|
### Prompt
Create a solution in CPP for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
int ara[n][n];
int i, j;
for (i = 0; i < n; i++) {
ara[0][i] = 1;
}
for (i = 1; i < n; i++) {
ara[i][0] = 1;
for (j = 1; j < n; j++) {
ara[i][j] = ara[i - 1][j] + ara[i][j - 1];
}
}
printf("%d", ara[n - 1][n - 1]);
}
```
|
#include <bits/stdc++.h>
bool isT = false;
void solv(int curt);
int main() {
std::ios::sync_with_stdio(0);
std::cin.tie(0);
if (isT) {
int t;
std::cin >> t;
for (int i = 1; i <= t; i++) {
solv(i);
}
return 0;
}
solv(0);
return 0;
}
int f(int r, int c) {
if (r == 1 || c == 1) return 1;
return f(r - 1, c) + f(r, c - 1);
}
void solv(int curt) {
int n;
std::cin >> n;
std::cout << f(n, n) << "\n";
}
|
### Prompt
In Cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
bool isT = false;
void solv(int curt);
int main() {
std::ios::sync_with_stdio(0);
std::cin.tie(0);
if (isT) {
int t;
std::cin >> t;
for (int i = 1; i <= t; i++) {
solv(i);
}
return 0;
}
solv(0);
return 0;
}
int f(int r, int c) {
if (r == 1 || c == 1) return 1;
return f(r - 1, c) + f(r, c - 1);
}
void solv(int curt) {
int n;
std::cin >> n;
std::cout << f(n, n) << "\n";
}
```
|
#include <bits/stdc++.h>
using namespace std;
bool isPowerOf2(long long int num) { return (num && (!(num & (num - 1)))); }
int main() {
int n = 0;
cin >> n;
std::vector<std::vector<int>> arr(n, std::vector<int>(n, 1));
int i = 0, j = 0, ma = 1;
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
ma = max(ma, arr[i][j]);
}
}
cout << ma << endl;
return 0;
}
|
### Prompt
In CPP, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool isPowerOf2(long long int num) { return (num && (!(num & (num - 1)))); }
int main() {
int n = 0;
cin >> n;
std::vector<std::vector<int>> arr(n, std::vector<int>(n, 1));
int i = 0, j = 0, ma = 1;
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
ma = max(ma, arr[i][j]);
}
}
cout << ma << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char* argv[]) {
int a;
cin >> a;
int** arr = (int**)malloc(sizeof(int*) * a);
for (int i = 0; i < a; i++) {
arr[i] = (int*)malloc(sizeof(int) * a);
}
for (int i = 0; i < a; i++) {
arr[0][i] = 1;
arr[i][0] = 1;
}
for (int i = 1; i < a; i++) {
for (int j = 1; j < a; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[a - 1][a - 1];
}
|
### Prompt
In Cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char* argv[]) {
int a;
cin >> a;
int** arr = (int**)malloc(sizeof(int*) * a);
for (int i = 0; i < a; i++) {
arr[i] = (int*)malloc(sizeof(int) * a);
}
for (int i = 0; i < a; i++) {
arr[0][i] = 1;
arr[i][0] = 1;
}
for (int i = 1; i < a; i++) {
for (int j = 1; j < a; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[a - 1][a - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long jkf(int n, int m) {
long long kk = 1;
for (int i = n; i > m; i--) {
kk *= i;
}
return kk;
}
int main() {
int a = 1, b;
cin >> a;
cout << jkf(2 * a - 2, a - 1) / jkf(a - 1, 1) << "\n";
}
|
### Prompt
Please provide a CPP coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long jkf(int n, int m) {
long long kk = 1;
for (int i = n; i > m; i--) {
kk *= i;
}
return kk;
}
int main() {
int a = 1, b;
cin >> a;
cout << jkf(2 * a - 2, a - 1) / jkf(a - 1, 1) << "\n";
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[10][10];
for (int i = 0; i < 10; i++) a[0][i] = 1;
for (int i = 1; i < 10; i++) a[i][0] = 1;
for (int i = 1; i < 10; i++) {
for (int j = 1; j < 10; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
int n;
cin >> n;
cout << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Please formulate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[10][10];
for (int i = 0; i < 10; i++) a[0][i] = 1;
for (int i = 1; i < 10; i++) a[i][0] = 1;
for (int i = 1; i < 10; i++) {
for (int j = 1; j < 10; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
int n;
cin >> n;
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, a[11][11];
int main() {
cin >> n;
a[0][1] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n][n];
return 0;
}
|
### Prompt
Your challenge is to write a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, a[11][11];
int main() {
cin >> n;
a[0][1] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n][n];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x;
cin >> x;
if (x == 1) {
cout << 1 << endl;
} else if (x == 2) {
cout << 2 << endl;
} else if (x == 3) {
cout << 6 << endl;
} else if (x == 4) {
cout << 20 << endl;
} else if (x == 5) {
cout << 70 << endl;
} else if (x == 6) {
cout << 252 << endl;
} else if (x == 7) {
cout << 924 << endl;
} else if (x == 8) {
cout << 3432 << endl;
} else if (x == 9) {
cout << 12870 << endl;
} else if (x == 10) {
cout << 48620 << endl;
}
return 0;
}
|
### Prompt
Generate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x;
cin >> x;
if (x == 1) {
cout << 1 << endl;
} else if (x == 2) {
cout << 2 << endl;
} else if (x == 3) {
cout << 6 << endl;
} else if (x == 4) {
cout << 20 << endl;
} else if (x == 5) {
cout << 70 << endl;
} else if (x == 6) {
cout << 252 << endl;
} else if (x == 7) {
cout << 924 << endl;
} else if (x == 8) {
cout << 3432 << endl;
} else if (x == 9) {
cout << 12870 << endl;
} else if (x == 10) {
cout << 48620 << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[11][11] = {1};
for (int i = 0; i < n; i++) {
a[i][0] = 1;
a[0][i] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << "\n" << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Please create a solution in cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[11][11] = {1};
for (int i = 0; i < n; i++) {
a[i][0] = 1;
a[0][i] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << "\n" << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) a[0][i] = 1;
for (int i = 0; i < n; i++) a[i][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << endl;
}
|
### Prompt
Construct a cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) a[0][i] = 1;
for (int i = 0; i < n; i++) a[i][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
}
|
### Prompt
Develop a solution in CPP to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << endl;
return 0;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[10][10], n;
int main() {
cin >> n;
for (int i = 0; i < 10; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < 10; i++)
for (int j = 1; j < 10; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
cout << a[n - 1][n - 1] << '\n';
return 0;
}
|
### Prompt
Develop a solution in cpp to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[10][10], n;
int main() {
cin >> n;
for (int i = 0; i < 10; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < 10; i++)
for (int j = 1; j < 10; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
cout << a[n - 1][n - 1] << '\n';
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int t, n, i, k, j, s = 0;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
continue;
}
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1];
}
|
### Prompt
Your task is to create a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t, n, i, k, j, s = 0;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
continue;
}
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int m[100][100];
int main(int argc, char const *argv[]) {
ios_base::sync_with_stdio(0);
long n, suma = 0;
cin >> n;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
m[i][j] = 1;
}
}
for (int i = 1; i < n; ++i) {
for (int j = 1; j < n; ++j) {
int x = m[i - 1][j], y = m[i][j - 1];
m[i][j] = x + y;
}
}
cout << m[n - 1][n - 1] << endl;
return 0;
}
|
### Prompt
Your challenge is to write a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int m[100][100];
int main(int argc, char const *argv[]) {
ios_base::sync_with_stdio(0);
long n, suma = 0;
cin >> n;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
m[i][j] = 1;
}
}
for (int i = 1; i < n; ++i) {
for (int j = 1; j < n; ++j) {
int x = m[i - 1][j], y = m[i][j - 1];
m[i][j] = x + y;
}
}
cout << m[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
inline T max(T &a, U &b) {
return a > b ? a : b;
}
template <class T, class U>
inline T min(T &a, U &b) {
return a < b ? a : b;
}
template <class T, class U>
inline T swap(T &a, U &b) {
T tmp = a;
a = b;
b = tmp;
}
template <class T>
T gcd(T a, T b) {
return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
return (a / gcd<T>(a, b) * b);
}
int LB(vector<int> v, int val) {
vector<int>::iterator low;
low = lower_bound(v.begin(), v.end(), val);
int ret = low - v.begin();
return ret;
}
int UB(vector<int> v, int val) {
vector<int>::iterator up;
up = upper_bound(v.begin(), v.end(), val);
int ret = up - v.begin();
return ret;
}
class CLASS_NAME {
public:
int method_NAME() {}
};
int arr[100][100];
int main() {
int n;
while (cin >> n) {
memset(arr, 0, sizeof(arr));
int m = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i == 1 || j == 1) arr[i][j] = 1;
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i == 1 || j == 1)
m = max(m, 1);
else {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
m = max(arr[i][j], m);
}
}
}
cout << m << endl;
}
return 0;
}
|
### Prompt
Generate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
inline T max(T &a, U &b) {
return a > b ? a : b;
}
template <class T, class U>
inline T min(T &a, U &b) {
return a < b ? a : b;
}
template <class T, class U>
inline T swap(T &a, U &b) {
T tmp = a;
a = b;
b = tmp;
}
template <class T>
T gcd(T a, T b) {
return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
return (a / gcd<T>(a, b) * b);
}
int LB(vector<int> v, int val) {
vector<int>::iterator low;
low = lower_bound(v.begin(), v.end(), val);
int ret = low - v.begin();
return ret;
}
int UB(vector<int> v, int val) {
vector<int>::iterator up;
up = upper_bound(v.begin(), v.end(), val);
int ret = up - v.begin();
return ret;
}
class CLASS_NAME {
public:
int method_NAME() {}
};
int arr[100][100];
int main() {
int n;
while (cin >> n) {
memset(arr, 0, sizeof(arr));
int m = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i == 1 || j == 1) arr[i][j] = 1;
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i == 1 || j == 1)
m = max(m, 1);
else {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
m = max(arr[i][j], m);
}
}
}
cout << m << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << endl;
}
|
### Prompt
Create a solution in cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[10][10];
int n, max = 0;
cin >> n;
for (int i = 0; i < n; i++) {
a[i][0] = 1;
a[0][i] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
max = a[0][0];
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
if (max < a[i][j]) max = a[i][j];
}
}
cout << max;
}
|
### Prompt
Please create a solution in Cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[10][10];
int n, max = 0;
cin >> n;
for (int i = 0; i < n; i++) {
a[i][0] = 1;
a[0][i] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
max = a[0][0];
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
if (max < a[i][j]) max = a[i][j];
}
}
cout << max;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j;
scanf("%d", &n);
int mat[n][n];
for (i = 0; i < n; ++i) mat[i][0] = mat[0][i] = 1;
for (i = 1; i < n; ++i)
for (j = 1; j < n; ++j) {
mat[i][j] = mat[i - 1][j] + mat[i][j - 1];
}
printf("%d", mat[n - 1][n - 1]);
return 0;
}
|
### Prompt
Please create a solution in CPP to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j;
scanf("%d", &n);
int mat[n][n];
for (i = 0; i < n; ++i) mat[i][0] = mat[0][i] = 1;
for (i = 1; i < n; ++i)
for (j = 1; j < n; ++j) {
mat[i][j] = mat[i - 1][j] + mat[i][j - 1];
}
printf("%d", mat[n - 1][n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const long long A = 100000000000000LL, N = 10000;
long long x[2], y[2], o[2], i, j, n, m;
int main() {
cin >> x[0] >> y[0] >> x[1] >> y[1];
o[0] = x[0] + y[0], o[1] = max(x[1], y[1]);
if (o[0] <= o[1])
cout << "Polycarp\n";
else if (x[0] <= x[1] && y[0] <= y[1])
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
}
|
### Prompt
Please provide a Cpp coded solution to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long A = 100000000000000LL, N = 10000;
long long x[2], y[2], o[2], i, j, n, m;
int main() {
cin >> x[0] >> y[0] >> x[1] >> y[1];
o[0] = x[0] + y[0], o[1] = max(x[1], y[1]);
if (o[0] <= o[1])
cout << "Polycarp\n";
else if (x[0] <= x[1] && y[0] <= y[1])
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 9;
const int mod = 1e9 + 7;
const int MOD = 1e9 + 696969;
const long long int INF = 1e18 + 3;
int xvas, yxas, xp, yp;
int main() {
cin >> xp >> yp >> xvas >> yxas;
int vas = 0;
int maxvas = max(xvas, yxas), minvas = min(xvas, yxas);
vas = minvas + maxvas - minvas;
int p = xp + yp;
if (xp <= xvas && yp <= yxas) {
cout << "Polycarp";
return 0;
}
if (p <= vas)
cout << "Polycarp";
else
cout << "Vasiliy";
}
|
### Prompt
Please create a solution in cpp to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 9;
const int mod = 1e9 + 7;
const int MOD = 1e9 + 696969;
const long long int INF = 1e18 + 3;
int xvas, yxas, xp, yp;
int main() {
cin >> xp >> yp >> xvas >> yxas;
int vas = 0;
int maxvas = max(xvas, yxas), minvas = min(xvas, yxas);
vas = minvas + maxvas - minvas;
int p = xp + yp;
if (xp <= xvas && yp <= yxas) {
cout << "Polycarp";
return 0;
}
if (p <= vas)
cout << "Polycarp";
else
cout << "Vasiliy";
}
```
|
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long powmod(long long a, long long b) {
long long res = 1;
a %= mod;
for (; b; b >>= 1) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
}
return res;
}
int xp, yp, xv, yv;
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xp <= xv && yp <= yv)
puts("Polycarp");
else {
if (xp + yp <= max(xv, yv))
puts("Polycarp");
else
puts("Vasiliy");
}
}
|
### Prompt
Please create a solution in Cpp to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long powmod(long long a, long long b) {
long long res = 1;
a %= mod;
for (; b; b >>= 1) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
}
return res;
}
int xp, yp, xv, yv;
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xp <= xv && yp <= yv)
puts("Polycarp");
else {
if (xp + yp <= max(xv, yv))
puts("Polycarp");
else
puts("Vasiliy");
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
const string player1 = "Polycarp";
const string player2 = "Vasiliy";
int main() {
int u, v, x, y;
cin >> x >> y >> u >> v;
if (u <= x && v <= y) {
cout << player2;
return 0;
}
for (int i = v, j = u; i >= max(v - u, 0); i--, j--)
if (j <= x && i <= y)
if (v - i >= abs(j - x) + abs(y - i)) {
cout << player1;
return 0;
}
if (u < v) {
for (int i = v - u; i >= 0; i--)
if (x + abs(y - i) <= u + v - u - i) {
cout << player1;
return 0;
}
} else {
for (int i = u - v; i >= 0; i--)
if (y + abs(x - i) <= v + u - v - i) {
cout << player1;
return 0;
}
}
cout << player2;
return 0;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const string player1 = "Polycarp";
const string player2 = "Vasiliy";
int main() {
int u, v, x, y;
cin >> x >> y >> u >> v;
if (u <= x && v <= y) {
cout << player2;
return 0;
}
for (int i = v, j = u; i >= max(v - u, 0); i--, j--)
if (j <= x && i <= y)
if (v - i >= abs(j - x) + abs(y - i)) {
cout << player1;
return 0;
}
if (u < v) {
for (int i = v - u; i >= 0; i--)
if (x + abs(y - i) <= u + v - u - i) {
cout << player1;
return 0;
}
} else {
for (int i = u - v; i >= 0; i--)
if (y + abs(x - i) <= v + u - v - i) {
cout << player1;
return 0;
}
}
cout << player2;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2) && (y1 <= y2)) {
cout << "Polycarp" << endl;
return 0;
}
if ((x1 >= x2) && (y1 >= y2)) {
cout << "Vasiliy" << endl;
return 0;
}
if (x1 + y1 <= max(x2, y2))
cout << " Polycarp " << endl;
else
cout << "Vasiliy" << endl;
return 0;
}
|
### Prompt
In Cpp, your task is to solve the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2) && (y1 <= y2)) {
cout << "Polycarp" << endl;
return 0;
}
if ((x1 >= x2) && (y1 >= y2)) {
cout << "Vasiliy" << endl;
return 0;
}
if (x1 + y1 <= max(x2, y2))
cout << " Polycarp " << endl;
else
cout << "Vasiliy" << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, eq;
map<pair<int, int>, int> mp;
int main() {
cin >> c >> d >> a >> b;
if (a == b) eq = 1;
int now = 1;
while (a != 0 || b != 0) {
mp[{a, b}] = now;
if (a > 0 && b > 0)
a--, b--;
else if (a > 0)
a--;
else
b--;
now++;
}
mp[{a, b}] = now;
now = 1;
int f = c, s = d;
while (c > 0 || d > 0) {
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
if (eq) {
if (c > d)
c--;
else
d--;
} else if (c > 0)
c--;
else
d--;
now++;
}
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
now = 1;
c = f, d = s;
while (c > 0 || d > 0) {
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
if (eq) {
if (c > d)
c--;
else
d--;
} else if (d > 0)
d--;
else
c--;
now++;
}
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
now = 1;
c = f, d = s;
while (c > 0 || d > 0) {
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
if (eq) {
if (c >= d)
c--;
else
d--;
} else if (c > 0)
c--;
else
d--;
now++;
}
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
now = 1;
c = f, d = s;
while (c > 0 || d > 0) {
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
if (eq) {
if (c >= d)
c--;
else
d--;
} else if (d > 0)
d--;
else
c--;
now++;
}
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
cout << "Vasiliy";
}
|
### Prompt
Generate a CPP solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, eq;
map<pair<int, int>, int> mp;
int main() {
cin >> c >> d >> a >> b;
if (a == b) eq = 1;
int now = 1;
while (a != 0 || b != 0) {
mp[{a, b}] = now;
if (a > 0 && b > 0)
a--, b--;
else if (a > 0)
a--;
else
b--;
now++;
}
mp[{a, b}] = now;
now = 1;
int f = c, s = d;
while (c > 0 || d > 0) {
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
if (eq) {
if (c > d)
c--;
else
d--;
} else if (c > 0)
c--;
else
d--;
now++;
}
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
now = 1;
c = f, d = s;
while (c > 0 || d > 0) {
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
if (eq) {
if (c > d)
c--;
else
d--;
} else if (d > 0)
d--;
else
c--;
now++;
}
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
now = 1;
c = f, d = s;
while (c > 0 || d > 0) {
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
if (eq) {
if (c >= d)
c--;
else
d--;
} else if (c > 0)
c--;
else
d--;
now++;
}
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
now = 1;
c = f, d = s;
while (c > 0 || d > 0) {
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
if (eq) {
if (c >= d)
c--;
else
d--;
} else if (d > 0)
d--;
else
c--;
now++;
}
if (mp[{c, d}] >= now) {
cout << "Polycarp";
return 0;
}
cout << "Vasiliy";
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n;
void f(char* s1, char* s2, char* a) {
int u = 0, d = 0, c = 0;
while (u < n && d < n) {
if (s1[u] == s2[d]) {
a[c++] = s1[u];
} else {
}
}
}
int main() {
int x1, y1, x2, y2;
while (cin >> x1 >> y1 >> x2 >> y2) {
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2))
puts("Polycarp");
else
puts("Vasiliy");
}
}
|
### Prompt
Your task is to create a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
void f(char* s1, char* s2, char* a) {
int u = 0, d = 0, c = 0;
while (u < n && d < n) {
if (s1[u] == s2[d]) {
a[c++] = s1[u];
} else {
}
}
}
int main() {
int x1, y1, x2, y2;
while (cin >> x1 >> y1 >> x2 >> y2) {
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2))
puts("Polycarp");
else
puts("Vasiliy");
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios ::sync_with_stdio(0);
cin.tie(0);
cout << (fixed) << setprecision(9);
long long xp, yp, xv, yv;
while (cin >> xp >> yp >> xv >> yv) {
if (xp + yp <= max(xv, yv))
cout << "Polycarp" << endl;
else if (xp <= xv && yp <= yv)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
}
|
### Prompt
Please provide a Cpp coded solution to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios ::sync_with_stdio(0);
cin.tie(0);
cout << (fixed) << setprecision(9);
long long xp, yp, xv, yv;
while (cin >> xp >> yp >> xv >> yv) {
if (xp + yp <= max(xv, yv))
cout << "Polycarp" << endl;
else if (xp <= xv && yp <= yv)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
}
```
|
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
void solve() {
int x[2], y[2];
cin >> x[0] >> y[0];
cin >> x[1] >> y[1];
if (x[0] + y[0] <= max(x[1], y[1]) || (x[0] <= x[1] && y[0] <= y[1])) {
cout << "Polycarp";
return;
}
cout << "Vasiliy";
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
solve();
return 0;
}
|
### Prompt
Develop a solution in CPP to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
void solve() {
int x[2], y[2];
cin >> x[0] >> y[0];
cin >> x[1] >> y[1];
if (x[0] + y[0] <= max(x[1], y[1]) || (x[0] <= x[1] && y[0] <= y[1])) {
cout << "Polycarp";
return;
}
cout << "Vasiliy";
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
solve();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
int xp, yp, xv, yv;
int solve() {
while (true) {
if (xp == 0 && yp == 0) return 1;
if (xp == 0)
yp--;
else if (yp == 0)
xp--;
else {
if (xp - yp > xv - yv)
xp--;
else
yp--;
}
if (xp == 0 && yp == 0) return 1;
if (xv == 0 && yv == 0) return 0;
if (xv == 0)
yv--;
else if (yv == 0)
xv--;
else {
if (xv - 1 != xp || yv - 1 != yp)
xv--, yv--;
else
--xv;
}
if (xv == 0 && yv == 0) return 0;
}
}
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
puts(solve() ? "Polycarp" : "Vasiliy");
return 0;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
int xp, yp, xv, yv;
int solve() {
while (true) {
if (xp == 0 && yp == 0) return 1;
if (xp == 0)
yp--;
else if (yp == 0)
xp--;
else {
if (xp - yp > xv - yv)
xp--;
else
yp--;
}
if (xp == 0 && yp == 0) return 1;
if (xv == 0 && yv == 0) return 0;
if (xv == 0)
yv--;
else if (yv == 0)
xv--;
else {
if (xv - 1 != xp || yv - 1 != yp)
xv--, yv--;
else
--xv;
}
if (xv == 0 && yv == 0) return 0;
}
}
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
puts(solve() ? "Polycarp" : "Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv;
while (scanf("%d %d %d %d", &xp, &yp, &xv, &yv) != EOF) {
if ((xp == 1 && yp == 0) || (yp == 1 && xp == 0)) {
printf("Polycarp\n");
} else if (xp <= xv && yp <= yv) {
printf("Polycarp\n");
} else if (xv <= xp && yv <= yp) {
printf("Vasiliy\n");
} else if (xp > xv) {
int quant = min(xv, yv - yp);
int xdest = xv - quant;
int ydest = yp;
int distP = xp - xdest;
int distV = quant + (yv - quant - yp);
if (distP <= distV) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else {
int quant = min(yv, xv - xp);
int xdest = xp;
int ydest = yv - quant;
int distP = yp - ydest;
int distV = quant + (xv - quant - xp);
if (distP <= distV) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
}
}
return 0;
}
|
### Prompt
Create a solution in cpp for the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv;
while (scanf("%d %d %d %d", &xp, &yp, &xv, &yv) != EOF) {
if ((xp == 1 && yp == 0) || (yp == 1 && xp == 0)) {
printf("Polycarp\n");
} else if (xp <= xv && yp <= yv) {
printf("Polycarp\n");
} else if (xv <= xp && yv <= yp) {
printf("Vasiliy\n");
} else if (xp > xv) {
int quant = min(xv, yv - yp);
int xdest = xv - quant;
int ydest = yp;
int distP = xp - xdest;
int distV = quant + (yv - quant - yp);
if (distP <= distV) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else {
int quant = min(yv, xv - xp);
int xdest = xp;
int ydest = yv - quant;
int distP = yp - ydest;
int distV = quant + (xv - quant - xp);
if (distP <= distV) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
}
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main() {
cin >> xp >> yp >> xv >> yv;
string ret[] = {"Polycarp", "Vasiliy"};
int p = xp + yp;
int v = xv + yv - min(xv, yv);
int p1 = xv, p2 = yv, r = 0;
while (p1 > 0 && p2 > 0) {
if ((p1 == xp && p2 <= yp) || (p2 == yp && p1 <= xp)) {
break;
}
p1--, p2--;
r++;
}
int mv = xp - p1 + yp - p2;
if (p <= v || mv <= r) {
cout << ret[0] << endl;
} else {
cout << ret[1] << endl;
}
return 0;
}
|
### Prompt
In Cpp, your task is to solve the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main() {
cin >> xp >> yp >> xv >> yv;
string ret[] = {"Polycarp", "Vasiliy"};
int p = xp + yp;
int v = xv + yv - min(xv, yv);
int p1 = xv, p2 = yv, r = 0;
while (p1 > 0 && p2 > 0) {
if ((p1 == xp && p2 <= yp) || (p2 == yp && p1 <= xp)) {
break;
}
p1--, p2--;
r++;
}
int mv = xp - p1 + yp - p2;
if (p <= v || mv <= r) {
cout << ret[0] << endl;
} else {
cout << ret[1] << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MOD = 1e9 + 7;
const int INF32 = 1 << 30;
const long long INF64 = 1LL << 60;
const long double pi = acos(-1);
long long gcd(long long a, long long b) { return (!b ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long mul(long long a, long long b) { return a * b % MOD; }
long long modpow(long long b, long long i) {
long long s = 1;
while (i) {
if (i % 2) s = (s * b) % MOD;
b = (b * b) % MOD;
i /= 2;
}
return s;
}
long long inv(long long a) { return modpow(a, MOD - 2); }
void solve() {
int xa, ya, xb, yb;
cin >> xa >> ya >> xb >> yb;
cout << (xa + ya <= max(xb, yb) || (xa <= xb && ya <= yb) ? "Polycarp"
: "Vasiliy");
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
|
### Prompt
Please create a solution in cpp to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MOD = 1e9 + 7;
const int INF32 = 1 << 30;
const long long INF64 = 1LL << 60;
const long double pi = acos(-1);
long long gcd(long long a, long long b) { return (!b ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long mul(long long a, long long b) { return a * b % MOD; }
long long modpow(long long b, long long i) {
long long s = 1;
while (i) {
if (i % 2) s = (s * b) % MOD;
b = (b * b) % MOD;
i /= 2;
}
return s;
}
long long inv(long long a) { return modpow(a, MOD - 2); }
void solve() {
int xa, ya, xb, yb;
cin >> xa >> ya >> xb >> yb;
cout << (xa + ya <= max(xb, yb) || (xa <= xb && ya <= yb) ? "Polycarp"
: "Vasiliy");
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
long long res = 1;
while (y > 0) {
if (y & 1) res = (res * x);
y = y >> 1;
x = (x * x);
}
return res;
}
long long ncr(long long n, long long r) {
if (n < r) return 0;
long long res = 1;
if (r > n - r) r = n - r;
for (long long i = 0; i < r; i++) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
long long gcd(long long a, long long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long max(long long a, long long b) {
long long ans = a > b ? a : b;
return ans;
}
long long min(long long a, long long b) {
long long ans = a < b ? a : b;
return ans;
}
clock_t time_p = clock();
void rtime() {
time_p = clock() - time_p;
cerr << "******************\nTime taken : "
<< (double)(time_p) / CLOCKS_PER_SEC << "\n";
}
signed main() {
{
long long x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (max(x2, y2) >= x1 + y1 || (x2 >= x1 && y2 >= y1))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
}
rtime();
return 0;
}
|
### Prompt
Please create a solution in CPP to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
long long res = 1;
while (y > 0) {
if (y & 1) res = (res * x);
y = y >> 1;
x = (x * x);
}
return res;
}
long long ncr(long long n, long long r) {
if (n < r) return 0;
long long res = 1;
if (r > n - r) r = n - r;
for (long long i = 0; i < r; i++) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
long long gcd(long long a, long long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long max(long long a, long long b) {
long long ans = a > b ? a : b;
return ans;
}
long long min(long long a, long long b) {
long long ans = a < b ? a : b;
return ans;
}
clock_t time_p = clock();
void rtime() {
time_p = clock() - time_p;
cerr << "******************\nTime taken : "
<< (double)(time_p) / CLOCKS_PER_SEC << "\n";
}
signed main() {
{
long long x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (max(x2, y2) >= x1 + y1 || (x2 >= x1 && y2 >= y1))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
}
rtime();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
bool win(int px, int py, int vx, int vy) {
for (int q = 0; q < max(vx, vy); ++q) {
int a = vx, b = vy;
int qw = min(min(a, b), q);
a -= qw;
b -= qw;
if (a == 0)
b -= (q - qw);
else
a -= (q - qw);
int tx = a - 1, ty = b - 1;
if (ty < 0) ty++;
if (tx < 0) tx++;
int mv = abs(px - tx) + abs(py - ty);
if (mv <= q + 1) return 1;
}
return 0;
}
int main() {
int px, py, vx, vy;
cin >> px >> py >> vx >> vy;
if (win(px, py, vx, vy))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool win(int px, int py, int vx, int vy) {
for (int q = 0; q < max(vx, vy); ++q) {
int a = vx, b = vy;
int qw = min(min(a, b), q);
a -= qw;
b -= qw;
if (a == 0)
b -= (q - qw);
else
a -= (q - qw);
int tx = a - 1, ty = b - 1;
if (ty < 0) ty++;
if (tx < 0) tx++;
int mv = abs(px - tx) + abs(py - ty);
if (mv <= q + 1) return 1;
}
return 0;
}
int main() {
int px, py, vx, vy;
cin >> px >> py >> vx >> vy;
if (win(px, py, vx, vy))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main() {
scanf("%d %d %d %d", &xp, &yp, &xv, &yv);
if (xp + yp <= max(xv, yv) || (xp <= xv && yp <= yv))
printf("Polycarp");
else
printf("Vasiliy");
return 0;
}
|
### Prompt
Generate a cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main() {
scanf("%d %d %d %d", &xp, &yp, &xv, &yv);
if (xp + yp <= max(xv, yv) || (xp <= xv && yp <= yv))
printf("Polycarp");
else
printf("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a1, b1, a2, b2;
int main() {
scanf("%d%d%d%d", &a1, &b1, &a2, &b2);
if (a1 <= a2 && b1 <= b2) {
printf("Polycarp\n");
return 0;
}
if (a1 >= a2 && b1 >= b2) {
printf("Vasiliy\n");
return 0;
}
int d1 = a1 + b1, d2 = max(a2, b2);
if (d1 <= d2) {
printf("Polycarp\n");
return 0;
} else {
printf("Vasiliy\n");
return 0;
}
return 0;
}
|
### Prompt
Develop a solution in Cpp to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a1, b1, a2, b2;
int main() {
scanf("%d%d%d%d", &a1, &b1, &a2, &b2);
if (a1 <= a2 && b1 <= b2) {
printf("Polycarp\n");
return 0;
}
if (a1 >= a2 && b1 >= b2) {
printf("Vasiliy\n");
return 0;
}
int d1 = a1 + b1, d2 = max(a2, b2);
if (d1 <= d2) {
printf("Polycarp\n");
return 0;
} else {
printf("Vasiliy\n");
return 0;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x2 < y2) {
swap(x1, y1);
swap(x2, y2);
}
bool flag = false;
if (x1 + y1 <= x2) flag = true;
if (y1 <= y2 && x1 <= x2) flag = true;
if (flag)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
|
### Prompt
Please create a solution in cpp to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x2 < y2) {
swap(x1, y1);
swap(x2, y2);
}
bool flag = false;
if (x1 + y1 <= x2) flag = true;
if (y1 <= y2 && x1 <= x2) flag = true;
if (flag)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (x1 - 1 < x2 && y1 < y2) {
puts("Polycarp");
return 0;
}
if (x1 < x2 && y1 - 1 < y2) {
puts("Polycarp");
return 0;
}
if (x1 + y1 <= max(x2, y2)) {
puts("Polycarp");
return 0;
}
puts("Vasiliy");
return 0;
}
|
### Prompt
Please formulate a CPP solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (x1 - 1 < x2 && y1 < y2) {
puts("Polycarp");
return 0;
}
if (x1 < x2 && y1 - 1 < y2) {
puts("Polycarp");
return 0;
}
if (x1 + y1 <= max(x2, y2)) {
puts("Polycarp");
return 0;
}
puts("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
while (cin >> x1 >> y1 >> x2 >> y2) {
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2)) {
cout << "Polycarp" << endl;
} else {
cout << "Vasiliy" << endl;
}
}
return 0;
}
|
### Prompt
Your challenge is to write a CPP solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
while (cin >> x1 >> y1 >> x2 >> y2) {
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2)) {
cout << "Polycarp" << endl;
} else {
cout << "Vasiliy" << endl;
}
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2))
printf("Polycarp\n");
else
printf("Vasiliy\n");
return 0;
}
|
### Prompt
Create a solution in cpp for the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2))
printf("Polycarp\n");
else
printf("Vasiliy\n");
return 0;
}
```
|
#include <bits/stdc++.h>
const double eps = 1e-8;
const int MAXN = (int)1e9 + 5;
using namespace std;
int main(int argc, char const *argv[]) {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2))
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
const double eps = 1e-8;
const int MAXN = (int)1e9 + 5;
using namespace std;
int main(int argc, char const *argv[]) {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2))
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2) {
if (y2 <= x2 - x1) {
if (y1 <= x2 - x1)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
} else {
if (y1 <= y2)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
} else if (y1 <= y2) {
if (x2 <= y2 - y1) {
if (x1 <= y2 - y1)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
} else {
if (x1 <= x2)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
} else {
cout << "Vasiliy" << endl;
}
}
|
### Prompt
Create a solution in CPP for the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2) {
if (y2 <= x2 - x1) {
if (y1 <= x2 - x1)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
} else {
if (y1 <= y2)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
} else if (y1 <= y2) {
if (x2 <= y2 - y1) {
if (x1 <= y2 - y1)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
} else {
if (x1 <= x2)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
} else {
cout << "Vasiliy" << endl;
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
void openFile() {
ios_base ::sync_with_stdio(false);
cin.tie(NULL);
}
const int maxN = 1e5 + 5;
const int maxM = 1e6 + 5;
const long long INF = 1e9 + 7;
int N;
int x, y, a, b;
void enter() { cin >> x >> y >> a >> b; }
void solve() {
int da = x + y;
int db = max(a, b);
if (da <= db) {
puts("Polycarp");
} else if (x <= a && y <= b) {
puts("Polycarp");
} else {
puts("Vasiliy");
}
}
int main() {
openFile();
enter();
solve();
return 0;
}
|
### Prompt
Please provide a Cpp coded solution to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void openFile() {
ios_base ::sync_with_stdio(false);
cin.tie(NULL);
}
const int maxN = 1e5 + 5;
const int maxM = 1e6 + 5;
const long long INF = 1e9 + 7;
int N;
int x, y, a, b;
void enter() { cin >> x >> y >> a >> b; }
void solve() {
int da = x + y;
int db = max(a, b);
if (da <= db) {
puts("Polycarp");
} else if (x <= a && y <= b) {
puts("Polycarp");
} else {
puts("Vasiliy");
}
}
int main() {
openFile();
enter();
solve();
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 (a <= c && b <= d) {
printf("Polycarp\n");
return 0;
}
if (a + b <= max(c, d)) {
printf("Polycarp\n");
return 0;
}
printf("Vasiliy\n");
return 0;
}
|
### Prompt
Create a solution in Cpp for the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### 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 (a <= c && b <= d) {
printf("Polycarp\n");
return 0;
}
if (a + b <= max(c, d)) {
printf("Polycarp\n");
return 0;
}
printf("Vasiliy\n");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if (c <= a && d <= b) {
cout << "Vasiliy";
return 0;
}
int tmp = a + b;
if (c < a && d < tmp) {
cout << "Vasiliy";
return 0;
}
if (d < b && c < tmp) {
cout << "Vasiliy";
return 0;
}
cout << "Polycarp";
return 0;
}
|
### Prompt
Your task is to create a CPP solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if (c <= a && d <= b) {
cout << "Vasiliy";
return 0;
}
int tmp = a + b;
if (c < a && d < tmp) {
cout << "Vasiliy";
return 0;
}
if (d < b && c < tmp) {
cout << "Vasiliy";
return 0;
}
cout << "Polycarp";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
void win1() {
cout << "Polycarp";
exit(0);
}
void win2() {
cout << "Vasiliy";
exit(0);
}
int main() {
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2)) win1();
if (x2 <= x1 && y2 <= y1) win2();
int len1 = max(abs(x2 - x1), y2);
int len2 = max(abs(y2 - y1), x2);
if (len1 < y1 || len2 < x1) win2();
win1();
return 0;
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void win1() {
cout << "Polycarp";
exit(0);
}
void win2() {
cout << "Vasiliy";
exit(0);
}
int main() {
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2)) win1();
if (x2 <= x1 && y2 <= y1) win2();
int len1 = max(abs(x2 - x1), y2);
int len2 = max(abs(y2 - y1), x2);
if (len1 < y1 || len2 < x1) win2();
win1();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int disv = max(x2, y2);
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp";
} else if (disv >= x1 + y1) {
cout << "Polycarp";
} else {
cout << "Vasiliy";
}
return 0;
}
|
### Prompt
Generate a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int disv = max(x2, y2);
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp";
} else if (disv >= x1 + y1) {
cout << "Polycarp";
} else {
cout << "Vasiliy";
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main(void) {
cin >> xp >> yp >> xv >> yv;
if (xv <= xp && yv <= yp) {
cout << "Vasiliy\n";
return 0;
}
if (xp <= xv && yp <= yv) {
cout << "Polycarp\n";
return 0;
}
int p = xp + yp;
int v = max(xv, yv);
cout << ((v >= p) ? "Polycarp\n" : "Vasiliy\n");
return 0;
}
|
### Prompt
Develop a solution in Cpp to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main(void) {
cin >> xp >> yp >> xv >> yv;
if (xv <= xp && yv <= yp) {
cout << "Vasiliy\n";
return 0;
}
if (xp <= xv && yp <= yv) {
cout << "Polycarp\n";
return 0;
}
int p = xp + yp;
int v = max(xv, yv);
cout << ((v >= p) ? "Polycarp\n" : "Vasiliy\n");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b;
cin >> a >> b;
int x, y;
cin >> x >> y;
bool win = true;
if (a <= x && b <= y) win = true;
if (x <= a && y <= b)
win = false;
else {
if (a > x) swap(a, b), swap(x, y);
int k = min(b, x - a);
b -= k;
x -= k;
y -= min(y, k);
win = (b <= y);
}
if (win)
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
|
### Prompt
Your task is to create a cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b;
cin >> a >> b;
int x, y;
cin >> x >> y;
bool win = true;
if (a <= x && b <= y) win = true;
if (x <= a && y <= b)
win = false;
else {
if (a > x) swap(a, b), swap(x, y);
int k = min(b, x - a);
b -= k;
x -= k;
y -= min(y, k);
win = (b <= y);
}
if (win)
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int N = 1000010, inf = 1000000007;
int a, b, x, y;
void fi() { printf("Polycarp"); }
void se() { printf("Vasiliy"); }
int main() {
scanf("%d %d %d %d", &a, &b, &x, &y);
if (a <= x && b <= y)
fi();
else if (y < b && x >= a + b)
fi();
else if (x < a && y >= a + b)
fi();
else
se();
return 0;
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1000010, inf = 1000000007;
int a, b, x, y;
void fi() { printf("Polycarp"); }
void se() { printf("Vasiliy"); }
int main() {
scanf("%d %d %d %d", &a, &b, &x, &y);
if (a <= x && b <= y)
fi();
else if (y < b && x >= a + b)
fi();
else if (x < a && y >= a + b)
fi();
else
se();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x, y, a, b;
scanf("%d %d %d %d", &x, &y, &a, &b);
if (x <= a && y <= b || x + y <= max(a, b)) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
return 0;
}
|
### Prompt
Create a solution in cpp for the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x, y, a, b;
scanf("%d %d %d %d", &x, &y, &a, &b);
if (x <= a && y <= b || x + y <= max(a, b)) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2 && y1 <= y2) || (x1 + y1 <= max(x2, y2)))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2 && y1 <= y2) || (x1 + y1 <= max(x2, y2)))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
inline int circRight(int a, unsigned m) {
return (a >> m) | (a << (CHAR_BIT - m));
}
inline int circLeft(int a, unsigned m) {
return (a << m) | (a >> (CHAR_BIT - m));
}
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int val = max(x2, y2);
if ((x1 + y1 <= val) || (x1 <= x2 && y1 <= y2)) {
cout << "Polycarp" << endl;
;
} else {
cout << "Vasiliy" << endl;
;
}
return 0;
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int circRight(int a, unsigned m) {
return (a >> m) | (a << (CHAR_BIT - m));
}
inline int circLeft(int a, unsigned m) {
return (a << m) | (a >> (CHAR_BIT - m));
}
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int val = max(x2, y2);
if ((x1 + y1 <= val) || (x1 <= x2 && y1 <= y2)) {
cout << "Polycarp" << endl;
;
} else {
cout << "Vasiliy" << endl;
;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (x1 <= x2 && y1 <= y2)
printf("Polycarp\n");
else if (x1 + y1 <= max(x2, y2))
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
|
### Prompt
Construct a cpp code solution to the problem outlined:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (x1 <= x2 && y1 <= y2)
printf("Polycarp\n");
else if (x1 + y1 <= max(x2, y2))
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
```
|
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
if (a < b) a = b;
}
inline void getint(int &first) {
char c;
while (c = getchar(), c > '9' || c < '0')
;
for (first = 0; c >= '0' && c <= '9'; c = getchar())
first = (first << 1) + (first << 3) + c - '0';
}
int power(int a, int b, int m, int ans = 1) {
for (; b; b >>= 1, a = 1LL * a * a % m)
if (b & 1) ans = 1LL * ans * a % m;
return ans;
}
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
int ok = 0;
if (xp + yp <= max(xv, yv))
ok = 1;
else if (xp + yp <= xv + yv) {
if (xp <= xv && yp <= yv)
ok = 1;
else if (xp <= xv && yp > yv) {
if (max(xv - xp, yp - yv) >= yp) ok = 1;
} else if (xp > xv && yp <= yv) {
if (max(yv - yp, xp - xv) >= xp) ok = 1;
}
}
puts(ok ? "Polycarp" : "Vasiliy");
return 0;
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
if (a < b) a = b;
}
inline void getint(int &first) {
char c;
while (c = getchar(), c > '9' || c < '0')
;
for (first = 0; c >= '0' && c <= '9'; c = getchar())
first = (first << 1) + (first << 3) + c - '0';
}
int power(int a, int b, int m, int ans = 1) {
for (; b; b >>= 1, a = 1LL * a * a % m)
if (b & 1) ans = 1LL * ans * a % m;
return ans;
}
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
int ok = 0;
if (xp + yp <= max(xv, yv))
ok = 1;
else if (xp + yp <= xv + yv) {
if (xp <= xv && yp <= yv)
ok = 1;
else if (xp <= xv && yp > yv) {
if (max(xv - xp, yp - yv) >= yp) ok = 1;
} else if (xp > xv && yp <= yv) {
if (max(yv - yp, xp - xv) >= xp) ok = 1;
}
}
puts(ok ? "Polycarp" : "Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int inf = 1e9 + 7;
int px, py, vx, vy, cnt;
void next() {
vx--;
vy--;
if (vx == -1) vx = 0;
if (vy == -1) vy = 0;
}
int main() {
cin >> px >> py >> vx >> vy;
if (vx <= px && vy <= py) {
puts("Vasiliy");
return 0;
}
while (!(vx <= px + 1 && vy <= py + 1)) {
next();
cnt++;
}
while (1) {
next();
if (abs(vx - px) + abs(vy - py) <= cnt + 1) {
puts("Polycarp");
return 0;
}
if (vx == 0 && vy == 0) break;
cnt++;
}
puts("Vasiliy");
}
|
### Prompt
Construct a cpp code solution to the problem outlined:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int inf = 1e9 + 7;
int px, py, vx, vy, cnt;
void next() {
vx--;
vy--;
if (vx == -1) vx = 0;
if (vy == -1) vy = 0;
}
int main() {
cin >> px >> py >> vx >> vy;
if (vx <= px && vy <= py) {
puts("Vasiliy");
return 0;
}
while (!(vx <= px + 1 && vy <= py + 1)) {
next();
cnt++;
}
while (1) {
next();
if (abs(vx - px) + abs(vy - py) <= cnt + 1) {
puts("Polycarp");
return 0;
}
if (vx == 0 && vy == 0) break;
cnt++;
}
puts("Vasiliy");
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
while (cin >> x1 >> y1 >> x2 >> y2) {
int ans = 0;
if (x1 + y1 <= max(x2, y2)) ans = 1;
if (x1 <= x2 && y1 <= y2) ans = 1;
if (ans)
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
}
return 0;
}
|
### Prompt
In CPP, your task is to solve the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
while (cin >> x1 >> y1 >> x2 >> y2) {
int ans = 0;
if (x1 + y1 <= max(x2, y2)) ans = 1;
if (x1 <= x2 && y1 <= y2) ans = 1;
if (ans)
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x, y, a, b;
cin >> x >> y >> a >> b;
if (x <= a && y <= b) {
cout << "Polycarp\n";
} else if (x + y <= max(a, b)) {
cout << "Polycarp\n";
} else {
cout << "Vasiliy\n";
}
return 0;
}
|
### Prompt
Your challenge is to write a cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x, y, a, b;
cin >> x >> y >> a >> b;
if (x <= a && y <= b) {
cout << "Polycarp\n";
} else if (x + y <= max(a, b)) {
cout << "Polycarp\n";
} else {
cout << "Vasiliy\n";
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
printf("%s\n", xp + yp <= max(xv, yv) || (xp <= xv && yp <= yv) ? "Polycarp"
: "Vasiliy");
}
|
### Prompt
Please create a solution in Cpp to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
printf("%s\n", xp + yp <= max(xv, yv) || (xp <= xv && yp <= yv) ? "Polycarp"
: "Vasiliy");
}
```
|
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2)) {
cout << "Polycarp";
return 0;
}
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp";
return 0;
}
cout << "Vasiliy";
}
|
### Prompt
In cpp, your task is to solve the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2)) {
cout << "Polycarp";
return 0;
}
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp";
return 0;
}
cout << "Vasiliy";
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2)) return cout << "Polycarp", 0;
if (x1 <= x2 && y1 <= y2)
return cout << "Polycarp", 0;
else
return cout << "Vasiliy", 0;
}
|
### Prompt
Please create a solution in cpp to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2)) return cout << "Polycarp", 0;
if (x1 <= x2 && y1 <= y2)
return cout << "Polycarp", 0;
else
return cout << "Vasiliy", 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const long long N = 5e5 + 19;
long long n, m;
vector<long long> adj[N];
long long color[N];
vector<pair<long long, long long> > edge;
bool dfs(long long v);
int32_t main() {
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
long long x, xx, y, yy;
cin >> x >> y >> xx >> yy;
if (yy >= x + y || xx >= y + x || (yy >= y && xx >= x))
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const long long N = 5e5 + 19;
long long n, m;
vector<long long> adj[N];
long long color[N];
vector<pair<long long, long long> > edge;
bool dfs(long long v);
int32_t main() {
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
long long x, xx, y, yy;
cin >> x >> y >> xx >> yy;
if (yy >= x + y || xx >= y + x || (yy >= y && xx >= x))
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int win = 0;
win = ((x1 <= x2 && y1 <= y2) | (x1 + y1 <= max(x2, y2)));
if (win)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
|
### Prompt
Please provide a Cpp coded solution to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int win = 0;
win = ((x1 <= x2 && y1 <= y2) | (x1 + y1 <= max(x2, y2)));
if (win)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char** argv) {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
while (1) {
if (xp == 0)
--yp;
else if (yp == 0)
--xp;
else {
int b = yv - xv;
int ny = xp + b;
if (ny < yp)
--yp;
else
--xp;
}
if (xp == 0 && yp == 0) {
cout << "Polycarp" << endl;
break;
}
if (xv == 0) {
--yv;
} else if (yv == 0) {
--xv;
} else {
if (xp != xv - 1 || yp != yv - 1) {
--xv;
--yv;
} else
--xv;
}
if (xv == 0 && yv == 0) {
cout << "Vasiliy" << endl;
break;
}
}
return 0;
}
|
### Prompt
Please create a solution in CPP to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char** argv) {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
while (1) {
if (xp == 0)
--yp;
else if (yp == 0)
--xp;
else {
int b = yv - xv;
int ny = xp + b;
if (ny < yp)
--yp;
else
--xp;
}
if (xp == 0 && yp == 0) {
cout << "Polycarp" << endl;
break;
}
if (xv == 0) {
--yv;
} else if (yv == 0) {
--xv;
} else {
if (xp != xv - 1 || yp != yv - 1) {
--xv;
--yv;
} else
--xv;
}
if (xv == 0 && yv == 0) {
cout << "Vasiliy" << endl;
break;
}
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long px, py, vx, vy;
cin >> px >> py >> vx >> vy;
if (px >= vx && py >= vy) {
cout << "Vasiliy" << endl;
return 0;
}
if (py > vy - vx + px) {
long long mx, my;
mx = px;
my = max(0ll, vy - vx + px);
long long tim = py - my;
vx = vx - tim;
if (vx < mx)
cout << "Vasiliy" << endl;
else
cout << "Polycarp" << endl;
} else if (py < vy - vx + px) {
long long mx, my;
my = py;
mx = max(0ll, py - vy + vx);
long long tim = px - mx;
vy = vy - tim;
if (vy < my)
cout << "Vasiliy" << endl;
else
cout << "Polycarp" << endl;
} else if (py == vy - vx + px) {
cout << "Polycarp" << endl;
}
}
|
### Prompt
In CPP, your task is to solve the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long px, py, vx, vy;
cin >> px >> py >> vx >> vy;
if (px >= vx && py >= vy) {
cout << "Vasiliy" << endl;
return 0;
}
if (py > vy - vx + px) {
long long mx, my;
mx = px;
my = max(0ll, vy - vx + px);
long long tim = py - my;
vx = vx - tim;
if (vx < mx)
cout << "Vasiliy" << endl;
else
cout << "Polycarp" << endl;
} else if (py < vy - vx + px) {
long long mx, my;
my = py;
mx = max(0ll, py - vy + vx);
long long tim = px - mx;
vy = vy - tim;
if (vy < my)
cout << "Vasiliy" << endl;
else
cout << "Polycarp" << endl;
} else if (py == vy - vx + px) {
cout << "Polycarp" << endl;
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int first, second, x2, y2;
int main() {
ios_base::sync_with_stdio(0);
string poly = "Polycarp\n", vas = "Vasiliy\n";
cin >> first >> second >> x2 >> y2;
if (first <= x2 && second <= y2) {
cout << poly;
return 0;
}
int a = first + second, b = max(x2, y2);
if (a <= b) {
cout << poly;
return 0;
}
cout << vas;
return 0;
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int first, second, x2, y2;
int main() {
ios_base::sync_with_stdio(0);
string poly = "Polycarp\n", vas = "Vasiliy\n";
cin >> first >> second >> x2 >> y2;
if (first <= x2 && second <= y2) {
cout << poly;
return 0;
}
int a = first + second, b = max(x2, y2);
if (a <= b) {
cout << poly;
return 0;
}
cout << vas;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, x, y, s1, s2;
cin >> a >> b >> x >> y;
s1 = a + b;
s2 = max(x, y);
if (s2 >= s1 || (a <= x && b <= y))
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
|
### Prompt
Develop a solution in cpp to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, x, y, s1, s2;
cin >> a >> b >> x >> y;
s1 = a + b;
s2 = max(x, y);
if (s2 >= s1 || (a <= x && b <= y))
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int x1, y1, x2, y2, m1, m2;
while (~scanf("%d%d%d%d", &x1, &y1, &x2, &y2)) {
if (x1 <= x2 && y1 <= y2)
puts("Polycarp");
else {
m1 = x1 + y1;
m2 = x2 > y2 ? x2 : y2;
if (m1 <= m2)
puts("Polycarp");
else
puts("Vasiliy");
}
}
return 0;
}
|
### Prompt
Construct a cpp code solution to the problem outlined:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int x1, y1, x2, y2, m1, m2;
while (~scanf("%d%d%d%d", &x1, &y1, &x2, &y2)) {
if (x1 <= x2 && y1 <= y2)
puts("Polycarp");
else {
m1 = x1 + y1;
m2 = x2 > y2 ? x2 : y2;
if (m1 <= m2)
puts("Polycarp");
else
puts("Vasiliy");
}
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T abs(T x) {
return x > 0 ? x : -x;
}
int pw() {
cout << "Polycarp";
return 0;
}
int vw() {
cout << "Vasiliy";
return 0;
}
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp + yp > xv + yv) {
return vw();
}
if (max(xv, yv) >= xp + yp) {
return pw();
}
if (xp <= xv && yp <= yv)
return pw();
else
return vw();
return 0;
}
|
### Prompt
Please create a solution in cpp to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T abs(T x) {
return x > 0 ? x : -x;
}
int pw() {
cout << "Polycarp";
return 0;
}
int vw() {
cout << "Vasiliy";
return 0;
}
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp + yp > xv + yv) {
return vw();
}
if (max(xv, yv) >= xp + yp) {
return pw();
}
if (xp <= xv && yp <= yv)
return pw();
else
return vw();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
bool polyw = false;
if ((xp == 0 && yp == 1) || (xp == 1 && yp == 0))
polyw = true;
else if (xp <= xv && yp <= yv)
polyw = true;
else if (xp + yp <= (xv > yv ? xv : yv))
polyw = true;
cout << (polyw ? "Polycarp" : "Vasiliy") << endl;
return 0;
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
bool polyw = false;
if ((xp == 0 && yp == 1) || (xp == 1 && yp == 0))
polyw = true;
else if (xp <= xv && yp <= yv)
polyw = true;
else if (xp + yp <= (xv > yv ? xv : yv))
polyw = true;
cout << (polyw ? "Polycarp" : "Vasiliy") << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a, b, c, d;
int main() {
cin >> a >> b >> c >> d;
if (a + b <= max(c, d) || (a <= c && b <= d))
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
|
### Prompt
Your task is to create a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a, b, c, d;
int main() {
cin >> a >> b >> c >> d;
if (a + b <= max(c, d) || (a <= c && b <= d))
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 >= x2 && y1 >= y2) {
cout << "Vasiliy\n";
exit(0);
}
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp\n";
exit(0);
}
if (x1 >= x2) {
int t2 = y2 - y1;
int t1 = x1 - (x2 - t2);
if (t1 <= t2) {
cout << "Polycarp\n";
exit(0);
}
}
if (y1 >= y2) {
int t2 = x2 - x1;
int t1 = y1 - (y2 - t2);
if (t1 <= t2) {
cout << "Polycarp\n";
exit(0);
}
}
int amt = min(x2, y2);
if (x1 + y1 <= (x2 - amt) + (y2 - amt) + amt) {
cout << "Polycarp\n";
exit(0);
}
cout << "Vasiliy\n";
exit(0);
}
|
### Prompt
Develop a solution in CPP to the problem described below:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 >= x2 && y1 >= y2) {
cout << "Vasiliy\n";
exit(0);
}
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp\n";
exit(0);
}
if (x1 >= x2) {
int t2 = y2 - y1;
int t1 = x1 - (x2 - t2);
if (t1 <= t2) {
cout << "Polycarp\n";
exit(0);
}
}
if (y1 >= y2) {
int t2 = x2 - x1;
int t1 = y1 - (y2 - t2);
if (t1 <= t2) {
cout << "Polycarp\n";
exit(0);
}
}
int amt = min(x2, y2);
if (x1 + y1 <= (x2 - amt) + (y2 - amt) + amt) {
cout << "Polycarp\n";
exit(0);
}
cout << "Vasiliy\n";
exit(0);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
srand(time(0));
int Xp, Yp, Xv, Yv;
cin >> Xp >> Yp >> Xv >> Yv;
if (Xp + Yp <= max(Xv, Yv)) {
cout << "Polycarp\n";
return 0;
}
if (Xp <= Xv && Yp <= Yv) {
cout << "Polycarp\n";
return 0;
}
cout << "Vasiliy\n";
return 0;
}
|
### Prompt
Please create a solution in cpp to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
srand(time(0));
int Xp, Yp, Xv, Yv;
cin >> Xp >> Yp >> Xv >> Yv;
if (Xp + Yp <= max(Xv, Yv)) {
cout << "Polycarp\n";
return 0;
}
if (Xp <= Xv && Yp <= Yv) {
cout << "Polycarp\n";
return 0;
}
cout << "Vasiliy\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int xp, yp, xc, yc;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> xp >> yp >> xc >> yc;
vector<pair<int, int>> store;
while (xc != 0 || yc != 0) {
store.push_back({xc, yc});
xc--;
yc--;
xc = max(0, xc);
yc = max(0, yc);
}
store.push_back({xc, yc});
for (int i = 0; i < store.size(); i++) {
pair<int, int> h = store[i];
int z = max(0, xp - h.first) + max(0, yp - h.second);
if (z <= i) {
cout << "Polycarp";
return 0;
}
}
cout << "Vasiliy";
return 0;
}
|
### Prompt
Please create a solution in CPP to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int xp, yp, xc, yc;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> xp >> yp >> xc >> yc;
vector<pair<int, int>> store;
while (xc != 0 || yc != 0) {
store.push_back({xc, yc});
xc--;
yc--;
xc = max(0, xc);
yc = max(0, yc);
}
store.push_back({xc, yc});
for (int i = 0; i < store.size(); i++) {
pair<int, int> h = store[i];
int z = max(0, xp - h.first) + max(0, yp - h.second);
if (z <= i) {
cout << "Polycarp";
return 0;
}
}
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long int pow_mod(long long int a, long long int b) {
long long int res = 1;
while (b) {
if (b & 1) res = (res * a) % 1000000007;
a = (a * a) % 1000000007;
b >>= 1;
}
return res;
}
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 && y1 <= y2)
cout << "Polycarp\n";
else if (x1 >= x2 && y1 >= y2)
cout << "Vasiliy\n";
else if (x1 + y1 <= max(x2, y2))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
|
### Prompt
Your task is to create a Cpp solution to the following problem:
Polycarp and Vasiliy love simple logical games. Today they play a game with infinite chessboard and one pawn for each player. Polycarp and Vasiliy move in turns, Polycarp starts. In each turn Polycarp can move his pawn from cell (x, y) to (x - 1, y) or (x, y - 1). Vasiliy can move his pawn from (x, y) to one of cells: (x - 1, y), (x - 1, y - 1) and (x, y - 1). Both players are also allowed to skip move.
There are some additional restrictions — a player is forbidden to move his pawn to a cell with negative x-coordinate or y-coordinate or to the cell containing opponent's pawn The winner is the first person to reach cell (0, 0).
You are given the starting coordinates of both pawns. Determine who will win if both of them play optimally well.
Input
The first line contains four integers: xp, yp, xv, yv (0 ≤ xp, yp, xv, yv ≤ 105) — Polycarp's and Vasiliy's starting coordinates.
It is guaranteed that in the beginning the pawns are in different cells and none of them is in the cell (0, 0).
Output
Output the name of the winner: "Polycarp" or "Vasiliy".
Examples
Input
2 1 2 2
Output
Polycarp
Input
4 7 7 4
Output
Vasiliy
Note
In the first sample test Polycarp starts in (2, 1) and will move to (1, 1) in the first turn. No matter what his opponent is doing, in the second turn Polycarp can move to (1, 0) and finally to (0, 0) in the third turn.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int pow_mod(long long int a, long long int b) {
long long int res = 1;
while (b) {
if (b & 1) res = (res * a) % 1000000007;
a = (a * a) % 1000000007;
b >>= 1;
}
return res;
}
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 && y1 <= y2)
cout << "Polycarp\n";
else if (x1 >= x2 && y1 >= y2)
cout << "Vasiliy\n";
else if (x1 + y1 <= max(x2, y2))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
```
|
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