output
stringlengths 52
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stringlengths 296
182k
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|---|---|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[2], b[2];
int num1, num2;
while (cin >> a[0] >> b[0] >> a[1] >> b[1]) {
if (a[0] <= a[1] && b[0] <= b[1]) {
puts("Polycarp");
continue;
}
num1 = a[0] + b[0];
num2 = max(a[1], b[1]);
puts(num1 <= num2 ? "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 main() {
int a[2], b[2];
int num1, num2;
while (cin >> a[0] >> b[0] >> a[1] >> b[1]) {
if (a[0] <= a[1] && b[0] <= b[1]) {
puts("Polycarp");
continue;
}
num1 = a[0] + b[0];
num2 = max(a[1], b[1]);
puts(num1 <= num2 ? "Polycarp" : "Vasiliy");
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int d2 = max(x2, y2);
int d1 = x1 + y1;
if ((x1 <= x2 && y1 <= y2) || d1 <= d2) {
cout << "Polycarp" << endl;
return 0;
}
{
cout << "Vasiliy" << endl;
return 0;
}
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;
cin >> x1 >> y1 >> x2 >> y2;
int d2 = max(x2, y2);
int d1 = x1 + y1;
if ((x1 <= x2 && y1 <= y2) || d1 <= d2) {
cout << "Polycarp" << endl;
return 0;
}
{
cout << "Vasiliy" << endl;
return 0;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= x2 + y2 - min(x2, y2))
cout << "Polycarp" << endl;
else if (x1 + y1 > x2 + y2)
cout << "Vasiliy" << endl;
else {
int d1 = x1 - y1;
int d2 = x2 - y2;
int dif = d1 - d2;
if (dif < 0) dif *= -1;
if (x1 + y1 <= x2 + y2 - min(min(x2, y2), max(0, dif - 1)))
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << 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() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= x2 + y2 - min(x2, y2))
cout << "Polycarp" << endl;
else if (x1 + y1 > x2 + y2)
cout << "Vasiliy" << endl;
else {
int d1 = x1 - y1;
int d2 = x2 - y2;
int dif = d1 - d2;
if (dif < 0) dif *= -1;
if (x1 + y1 <= x2 + y2 - min(min(x2, y2), max(0, dif - 1)))
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int N = 252;
int dis(int x1, int y1, int x2, int y2) { return abs(x1 - x2) + abs(y1 - y2); }
bool Gao(int x1, int y1, int x2, int y2) {
if (dis(0, 0, x1, y1) <= max(x2, y2)) return true;
for (int i = 1; i <= min(x2, y2); ++i) {
if (dis(x2 - i, y2 - i, x1, y1) <= i) return true;
}
return false;
}
int main() {
int x1, x2, y1, y2;
while (cin >> x1 >> y1 >> x2 >> y2) {
if (Gao(x1, y1, x2, y2))
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
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 = 252;
int dis(int x1, int y1, int x2, int y2) { return abs(x1 - x2) + abs(y1 - y2); }
bool Gao(int x1, int y1, int x2, int y2) {
if (dis(0, 0, x1, y1) <= max(x2, y2)) return true;
for (int i = 1; i <= min(x2, y2); ++i) {
if (dis(x2 - i, y2 - i, x1, y1) <= i) return true;
}
return false;
}
int main() {
int x1, x2, y1, y2;
while (cin >> x1 >> y1 >> x2 >> y2) {
if (Gao(x1, y1, x2, y2))
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
FILE *in;
FILE *out;
const int MAX = 1024;
int main(void) {
in = stdin;
out = stdout;
int x1, y1, x2, y2;
fscanf(in, "%d %d %d %d", &x1, &y1, &x2, &y2);
bool canWin = false;
if (x1 <= x2 && y1 <= y2) canWin = true;
if (x1 + y1 <= max(x2, y2)) canWin = true;
fprintf(out, "%s\n", canWin ? "Polycarp" : "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;
FILE *in;
FILE *out;
const int MAX = 1024;
int main(void) {
in = stdin;
out = stdout;
int x1, y1, x2, y2;
fscanf(in, "%d %d %d %d", &x1, &y1, &x2, &y2);
bool canWin = false;
if (x1 <= x2 && y1 <= y2) canWin = true;
if (x1 + y1 <= max(x2, y2)) canWin = true;
fprintf(out, "%s\n", canWin ? "Polycarp" : "Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
const int MAXE = 200005;
const long long INF = 1e18;
struct Edge {
int v, n;
Edge() {}
Edge(int v, int n) : v(v), n(n) {}
};
Edge E[MAXE];
int H[MAXN], cntE;
int val[MAXN];
long long dp[MAXN][2];
int n;
long long ans;
void clear() {
cntE = 0;
memset(H, -1, sizeof H);
}
void addedge(int u, int v) {
E[cntE] = Edge(v, H[u]);
H[u] = cntE++;
}
long long dfs(int u) {
long long minv = INF;
long long sum = 0;
int cnt = 0;
dp[u][0] = 0;
dp[u][1] = val[u];
for (int i = H[u]; ~i; i = E[i].n) {
int v = E[i].v;
dfs(v);
long long tmp = max(dp[v][1], dp[v][0]);
minv = dp[u][0] += tmp;
dp[u][1] += tmp;
minv = min(minv, tmp + val[v]);
++cnt;
}
if (cnt & 1) sum -= minv;
ans = max(ans, sum + val[u]);
return sum;
}
int a, b, c, d;
void solve() {
if (a + b <= min(c, d) + max(c, d) - min(c, d)) {
printf("Polycarp\n");
return;
}
if (a > c || b > d)
printf("Vasiliy\n");
else
printf("Polycarp\n");
}
int main() {
while (~scanf("%d%d%d%d", &a, &b, &c, &d)) solve();
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;
const int MAXN = 200005;
const int MAXE = 200005;
const long long INF = 1e18;
struct Edge {
int v, n;
Edge() {}
Edge(int v, int n) : v(v), n(n) {}
};
Edge E[MAXE];
int H[MAXN], cntE;
int val[MAXN];
long long dp[MAXN][2];
int n;
long long ans;
void clear() {
cntE = 0;
memset(H, -1, sizeof H);
}
void addedge(int u, int v) {
E[cntE] = Edge(v, H[u]);
H[u] = cntE++;
}
long long dfs(int u) {
long long minv = INF;
long long sum = 0;
int cnt = 0;
dp[u][0] = 0;
dp[u][1] = val[u];
for (int i = H[u]; ~i; i = E[i].n) {
int v = E[i].v;
dfs(v);
long long tmp = max(dp[v][1], dp[v][0]);
minv = dp[u][0] += tmp;
dp[u][1] += tmp;
minv = min(minv, tmp + val[v]);
++cnt;
}
if (cnt & 1) sum -= minv;
ans = max(ans, sum + val[u]);
return sum;
}
int a, b, c, d;
void solve() {
if (a + b <= min(c, d) + max(c, d) - min(c, d)) {
printf("Polycarp\n");
return;
}
if (a > c || b > d)
printf("Vasiliy\n");
else
printf("Polycarp\n");
}
int main() {
while (~scanf("%d%d%d%d", &a, &b, &c, &d)) solve();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
void swap(int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
long long int power(long long int a, long long int b) {
long long int x = 1;
long long int y = a;
while (b > 0) {
if (b & 1) {
x = x * y;
x %= 1000000007;
}
y = y * y;
y %= 1000000007;
b /= 2;
}
return x;
}
long long int inver(long long int a) { return power(a, 1000000007 - 2); }
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int gcd(int a, int b) {
while (b != 0) {
int temp = a;
a = b;
b = temp % a;
}
return a;
}
int innumber(char c) { return c - '0'; }
bool isprime(long long int x) {
for (long long int i = 2; i <= sqrt(x); i++) {
if (x % i == 0) return false;
}
return true;
}
int main() {
int a, b, c, d;
ios_base::sync_with_stdio(0);
cin >> a >> b >> c >> d;
if (a + b <= max(c, d)) {
cout << "Polycarp" << endl;
return 0;
} else if (a <= c && b <= d) {
cout << "Polycarp" << endl;
return 0;
} else {
cout << "Vasiliy" << endl;
return 0;
}
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;
void swap(int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
long long int power(long long int a, long long int b) {
long long int x = 1;
long long int y = a;
while (b > 0) {
if (b & 1) {
x = x * y;
x %= 1000000007;
}
y = y * y;
y %= 1000000007;
b /= 2;
}
return x;
}
long long int inver(long long int a) { return power(a, 1000000007 - 2); }
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int gcd(int a, int b) {
while (b != 0) {
int temp = a;
a = b;
b = temp % a;
}
return a;
}
int innumber(char c) { return c - '0'; }
bool isprime(long long int x) {
for (long long int i = 2; i <= sqrt(x); i++) {
if (x % i == 0) return false;
}
return true;
}
int main() {
int a, b, c, d;
ios_base::sync_with_stdio(0);
cin >> a >> b >> c >> d;
if (a + b <= max(c, d)) {
cout << "Polycarp" << endl;
return 0;
} else if (a <= c && b <= d) {
cout << "Polycarp" << endl;
return 0;
} else {
cout << "Vasiliy" << endl;
return 0;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
long long r;
while (b != 0) {
r = a % b;
a = b;
b = r;
}
return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 100010;
int xp, yp;
int xv, yv;
void solve() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xp + yp <= max(xv, yv) || (xp <= xv && yp <= yv))
printf("Polycarp");
else
printf("Vasiliy");
}
int main() {
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;
long long gcd(long long a, long long b) {
long long r;
while (b != 0) {
r = a % b;
a = b;
b = r;
}
return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 100010;
int xp, yp;
int xv, yv;
void solve() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xp + yp <= max(xv, yv) || (xp <= xv && yp <= yv))
printf("Polycarp");
else
printf("Vasiliy");
}
int main() {
solve();
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);
int flag = 0;
int s1 = x1 + y1;
int s2 = max(x2, y2);
if (s1 <= s2) flag = 1;
if (x1 >= 0 && x1 <= x2 && y1 >= 0 && y1 <= y2) flag = 1;
if (flag)
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
|
### 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;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int flag = 0;
int s1 = x1 + y1;
int s2 = max(x2, y2);
if (s1 <= s2) flag = 1;
if (x1 >= 0 && x1 <= x2 && y1 >= 0 && y1 <= y2) flag = 1;
if (flag)
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long mod = (long long)1 << 62;
long long QPow(long long a, long long b) {
long long ans = 1;
while (b > 0) {
if (b & 1) ans = ans % mod * a % mod;
a = a % mod * a % mod;
b >>= 1;
}
return ans;
}
long long inf = (long long)1 << 60;
double eps = 1e-8;
long long fac[22];
long long fac_1[22];
long long inv[22];
void getInv() {
inv[1] = 1;
for (int i = 2; i <= 200001; i++)
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
void init1() {
fac[0] = 1;
fac_1[0] = 1;
fac_1[1] = 1;
for (int i = 1; i <= 200001; i++)
fac[i] = fac[i - 1] * i % mod, fac_1[i] = inv[i] * fac_1[i - 1] % mod;
}
long long C(int n, int m) {
if (m < 0 || m > n || n < 0) return 0;
return fac[n] * fac_1[n - m] % mod * fac_1[m] % mod;
}
long long gcd(long long x, long long y) {
if (x == 1 || y == 1)
return 1;
else {
if (x == 0) return y;
if (y == 0) return x;
if (x % y == 0) return y;
if (y % x == 0) return x;
if (x > y)
return gcd(x % y, y);
else
return gcd(x, y % x);
}
}
int sign(long long x) {
if (x == 0) return 0;
if (x > 0) return 1;
return -1;
}
long long toInt(string str) {
long long ans = 0;
for (int i = (str[0] == '-'); i < str.size(); i++) {
ans *= 10;
ans += str[i] - '0';
}
if (str[0] == '-') ans = -ans;
return ans;
}
string toString(long long a) {
if (a == 0) return "0";
string temp = "";
if (a < 0) temp.push_back('-');
long long tp = a;
while (tp) {
temp.push_back('0' + tp % 10);
tp /= 10;
}
string temp2 = temp;
for (int ii = 0; ii < temp.size(); ii++) {
temp[ii] = temp2[temp.size() - 1 - ii];
}
return temp;
}
const int maxn = 105;
bool equ(string s, string t) {
if (s.size() == 1) return s[0] == t[0];
bool flag = 1;
for (int i = 0; i < s.size(); i++)
if (s[i] != t[i]) {
flag = 0;
break;
}
if (flag) return 1;
if (s.size() % 2 != 0) return 0;
string s1, s2, t1, t2;
string t3, t4;
int slen = s.size(), tlen = t.size();
for (int i = 0; i < slen / 2; i++) s1.push_back(s[i]);
for (int i = slen / 2; i < slen; i++) s2.push_back(s[i]);
for (int i = 0; i < tlen / 2; i++) t1.push_back(t[i]);
for (int i = tlen / 2; i < tlen; i++) t2.push_back(t[i]);
for (int i = 0; i < tlen / 2; i++) t3.push_back(t[tlen - 1 - i]);
for (int i = tlen / 2; i < tlen; i++) t4.push_back(t[tlen - 1 - i]);
flag |= equ(s1, t1) && equ(s2, t2);
flag |= equ(s1, t3) && equ(s2, t4);
return flag;
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
ifstream in("Text.txt");
long long n, x, t, m, k, d, L;
long long xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
bool win = 1;
if (xp >= xv && yp >= yv) {
win = 0;
} else if (xp <= xv && yp <= yv)
win = 1;
else if (xv < yv) {
if (yp > yv)
win = 0;
else
win = xp <= yv - yp;
} else if (xv > yv) {
if (yp < yv)
win = 0;
else
win = yp <= xv - xp;
} else
win = 0;
if (win)
cout << "Polycarp";
else
cout << "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;
long long mod = (long long)1 << 62;
long long QPow(long long a, long long b) {
long long ans = 1;
while (b > 0) {
if (b & 1) ans = ans % mod * a % mod;
a = a % mod * a % mod;
b >>= 1;
}
return ans;
}
long long inf = (long long)1 << 60;
double eps = 1e-8;
long long fac[22];
long long fac_1[22];
long long inv[22];
void getInv() {
inv[1] = 1;
for (int i = 2; i <= 200001; i++)
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
void init1() {
fac[0] = 1;
fac_1[0] = 1;
fac_1[1] = 1;
for (int i = 1; i <= 200001; i++)
fac[i] = fac[i - 1] * i % mod, fac_1[i] = inv[i] * fac_1[i - 1] % mod;
}
long long C(int n, int m) {
if (m < 0 || m > n || n < 0) return 0;
return fac[n] * fac_1[n - m] % mod * fac_1[m] % mod;
}
long long gcd(long long x, long long y) {
if (x == 1 || y == 1)
return 1;
else {
if (x == 0) return y;
if (y == 0) return x;
if (x % y == 0) return y;
if (y % x == 0) return x;
if (x > y)
return gcd(x % y, y);
else
return gcd(x, y % x);
}
}
int sign(long long x) {
if (x == 0) return 0;
if (x > 0) return 1;
return -1;
}
long long toInt(string str) {
long long ans = 0;
for (int i = (str[0] == '-'); i < str.size(); i++) {
ans *= 10;
ans += str[i] - '0';
}
if (str[0] == '-') ans = -ans;
return ans;
}
string toString(long long a) {
if (a == 0) return "0";
string temp = "";
if (a < 0) temp.push_back('-');
long long tp = a;
while (tp) {
temp.push_back('0' + tp % 10);
tp /= 10;
}
string temp2 = temp;
for (int ii = 0; ii < temp.size(); ii++) {
temp[ii] = temp2[temp.size() - 1 - ii];
}
return temp;
}
const int maxn = 105;
bool equ(string s, string t) {
if (s.size() == 1) return s[0] == t[0];
bool flag = 1;
for (int i = 0; i < s.size(); i++)
if (s[i] != t[i]) {
flag = 0;
break;
}
if (flag) return 1;
if (s.size() % 2 != 0) return 0;
string s1, s2, t1, t2;
string t3, t4;
int slen = s.size(), tlen = t.size();
for (int i = 0; i < slen / 2; i++) s1.push_back(s[i]);
for (int i = slen / 2; i < slen; i++) s2.push_back(s[i]);
for (int i = 0; i < tlen / 2; i++) t1.push_back(t[i]);
for (int i = tlen / 2; i < tlen; i++) t2.push_back(t[i]);
for (int i = 0; i < tlen / 2; i++) t3.push_back(t[tlen - 1 - i]);
for (int i = tlen / 2; i < tlen; i++) t4.push_back(t[tlen - 1 - i]);
flag |= equ(s1, t1) && equ(s2, t2);
flag |= equ(s1, t3) && equ(s2, t4);
return flag;
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
ifstream in("Text.txt");
long long n, x, t, m, k, d, L;
long long xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
bool win = 1;
if (xp >= xv && yp >= yv) {
win = 0;
} else if (xp <= xv && yp <= yv)
win = 1;
else if (xv < yv) {
if (yp > yv)
win = 0;
else
win = xp <= yv - yp;
} else if (xv > yv) {
if (yp < yv)
win = 0;
else
win = yp <= xv - xp;
} else
win = 0;
if (win)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int MAXN = -1;
struct dot {
long long x, y;
};
int n;
dot G, B;
int main() {
while (scanf("%lld %lld %lld %lld", &G.x, &G.y, &B.x, &B.y) == 4) {
if (B.x > B.y) {
swap(B.x, B.y);
swap(G.x, G.y);
}
if (G.x <= B.x && G.y <= B.y) {
printf("Polycarp\n");
continue;
}
cout << ((G.x > B.x) && (G.x + G.y <= B.y) ? "Polycarp" : "Vasiliy")
<< endl;
}
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;
const int MAXN = -1;
struct dot {
long long x, y;
};
int n;
dot G, B;
int main() {
while (scanf("%lld %lld %lld %lld", &G.x, &G.y, &B.x, &B.y) == 4) {
if (B.x > B.y) {
swap(B.x, B.y);
swap(G.x, G.y);
}
if (G.x <= B.x && G.y <= B.y) {
printf("Polycarp\n");
continue;
}
cout << ((G.x > B.x) && (G.x + G.y <= B.y) ? "Polycarp" : "Vasiliy")
<< endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
void wczytaj_dane(int x[], int y[]) {
for (int i = 0; i < (2); i++) cin >> x[i] >> y[i];
}
bool pomiedzy(int x, int a, int b) { return a <= x && x <= b; }
bool kwadrat(int x[], int y[]) {
int mini = min(x[1], y[1]);
int xx = x[1] - mini, yy = y[1] - mini;
return pomiedzy(x[0], xx, x[1]) && pomiedzy(y[0], yy, y[1]);
}
int odleglosc0(int x, int y) { return x + y; }
int odleglosc1(int x, int y) { return max(x, y); }
int rozwiaz(int x[], int y[]) {
if (kwadrat(x, y)) return 0;
return (odleglosc0(x[0], y[0]) <= odleglosc1(x[1], y[1]) ? 0 : 1);
}
void zrob_test() {
int x[2], y[2];
wczytaj_dane(x, y);
cout << (rozwiaz(x, y) == 0 ? "Polycarp\n" : "Vasiliy\n") << '\n';
}
int main() {
ios_base::sync_with_stdio(0);
zrob_test();
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;
void wczytaj_dane(int x[], int y[]) {
for (int i = 0; i < (2); i++) cin >> x[i] >> y[i];
}
bool pomiedzy(int x, int a, int b) { return a <= x && x <= b; }
bool kwadrat(int x[], int y[]) {
int mini = min(x[1], y[1]);
int xx = x[1] - mini, yy = y[1] - mini;
return pomiedzy(x[0], xx, x[1]) && pomiedzy(y[0], yy, y[1]);
}
int odleglosc0(int x, int y) { return x + y; }
int odleglosc1(int x, int y) { return max(x, y); }
int rozwiaz(int x[], int y[]) {
if (kwadrat(x, y)) return 0;
return (odleglosc0(x[0], y[0]) <= odleglosc1(x[1], y[1]) ? 0 : 1);
}
void zrob_test() {
int x[2], y[2];
wczytaj_dane(x, y);
cout << (rozwiaz(x, y) == 0 ? "Polycarp\n" : "Vasiliy\n") << '\n';
}
int main() {
ios_base::sync_with_stdio(0);
zrob_test();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int N, M;
int xx1, yy1, xx2, yy2;
int main() {
int i, j;
while (~scanf("%d%d%d%d", &xx1, &yy1, &xx2, &yy2)) {
if (xx1 <= xx2 && yy1 <= yy2 || xx1 + yy1 <= max(xx2, yy2)) {
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;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int N, M;
int xx1, yy1, xx2, yy2;
int main() {
int i, j;
while (~scanf("%d%d%d%d", &xx1, &yy1, &xx2, &yy2)) {
if (xx1 <= xx2 && yy1 <= yy2 || xx1 + yy1 <= max(xx2, yy2)) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int range(int x1, int y1, int x2, int y2) {
return abs(x1 - x2) + abs(y1 - y2);
}
int range2(int x1, int y1, int x2, int y2) {
if (abs(x1 - x2) <= abs(y1 - y2))
return abs(y1 - y2);
else
return abs(x1 - x2);
}
int main() {
int x1, y1, x2, y2;
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
if ((x1 <= x2 && y1 <= y2) || (x1 >= x2 && y1 >= y2)) {
if (x1 <= x2 && y1 <= y2)
printf("Polycarp");
else
printf("Vasiliy");
return 0;
}
if (x2 < x1) {
if (range(x1, y1, 0, y1) <= range2(x2, y2, 0, y1))
printf("Polycarp");
else
printf("Vasiliy");
} else {
if (range(x1, y1, x1, 0) <= range2(x2, y2, x1, 0))
printf("Polycarp");
else
printf("Vasiliy");
}
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 range(int x1, int y1, int x2, int y2) {
return abs(x1 - x2) + abs(y1 - y2);
}
int range2(int x1, int y1, int x2, int y2) {
if (abs(x1 - x2) <= abs(y1 - y2))
return abs(y1 - y2);
else
return abs(x1 - x2);
}
int main() {
int x1, y1, x2, y2;
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
if ((x1 <= x2 && y1 <= y2) || (x1 >= x2 && y1 >= y2)) {
if (x1 <= x2 && y1 <= y2)
printf("Polycarp");
else
printf("Vasiliy");
return 0;
}
if (x2 < x1) {
if (range(x1, y1, 0, y1) <= range2(x2, y2, 0, y1))
printf("Polycarp");
else
printf("Vasiliy");
} else {
if (range(x1, y1, x1, 0) <= range2(x2, y2, x1, 0))
printf("Polycarp");
else
printf("Vasiliy");
}
return 0;
}
```
|
#include <bits/stdc++.h>
#pragma gcc optimize("O3")
#pragma gcc target("sse4")
using namespace std;
mt19937 rng(2391);
template <typename A>
istream& operator>>(istream& fin, vector<A>& v) {
for (auto it = v.begin(); it != v.end(); ++it) fin >> *it;
return fin;
}
template <typename A, typename B>
istream& operator>>(istream& fin, pair<A, B>& p) {
fin >> p.first >> p.second;
return fin;
}
template <typename A, typename B>
pair<A, B> operator+(const pair<A, B>& a, const pair<A, B>& b) {
return make_pair(a.first + b.first, a.second + b.second);
}
template <typename A, typename B>
pair<A, B> operator+=(pair<A, B>& a, const pair<A, B>& b) {
a.first += b.first;
a.second += b.second;
return a;
}
template <typename A, typename B>
pair<A, B> operator-(const pair<A, B>& a, const pair<A, B>& b) {
return make_pair(a.first - b.first, a.second - b.second);
}
template <typename A, typename B>
pair<A, B> operator-(const pair<A, B>& a) {
return make_pair(-a.first, -a.second);
}
template <typename A, typename B>
pair<A, B>& operator++(pair<A, B>& a) {
++a.first;
++a.second;
return a;
}
template <typename A, typename B>
pair<A, B>& operator--(pair<A, B>& a) {
--a.first;
--a.second;
return a;
}
template <typename A>
vector<A>& operator++(vector<A>& a) {
for (auto it = a.begin(); it != a.end(); ++it) ++*it;
return a;
}
template <typename A>
vector<A>& operator--(vector<A>& a) {
for (auto it = a.begin(); it != a.end(); ++it) --*it;
return a;
}
template <typename A, typename B>
pair<A, B> operator++(pair<A, B>& a, int) {
pair<A, B> b = a;
++a;
return b;
}
template <typename A, typename B>
pair<A, B> operator--(pair<A, B>& a, int) {
pair<A, B> b = a;
--a;
return b;
}
template <typename A>
vector<A>& operator++(vector<A>& a, int) {
vector<A> b = a;
++a;
return b;
}
template <typename A>
vector<A>& operator--(vector<A>& a, int) {
vector<A> b = a;
--a;
return b;
}
vector<vector<int>> adjlist_from_edgelist(const vector<pair<int, int>>& e,
const int& n) {
vector<vector<int>> g(n);
for (auto it = e.begin(); it != e.end(); ++it) {
g[it->first].push_back(it->second);
g[it->second].push_back(it->first);
}
return g;
}
template <typename A, typename B>
pair<A, B> operator-=(pair<A, B>& a, const pair<A, B>& b) {
a.first -= b.first;
a.second -= b.second;
return a;
}
template <typename A>
A operator*(const pair<A, A>& p, const pair<A, A>& q) {
return p.first * q.first + p.second * q.second;
}
template <typename A>
pair<A, A> operator*(const pair<A, A>& p, const A& q) {
return make_pair(p.first * q, p.second * q);
}
template <typename A>
A operator%(const pair<A, A>& p, const pair<A, A>& q) {
return p.first * q.second - p.second * q.first;
}
template <typename A>
A sq_len(const pair<A, A>& p) {
return p * p;
}
template <typename A>
A sq_dist(const pair<A, A>& p, const pair<A, A>& q) {
return sq_len(p - q);
}
template <typename A>
double len(const pair<A, A>& p) {
return sqrt(sq_len(p));
}
template <typename A>
double dist(const pair<A, A>& p, const pair<A, A>& q) {
return len(p - q);
}
template <typename A>
bool is_rhombus(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c,
const pair<A, A>& d) {
A x = sq_dist(a, b);
A y = sq_dist(b, c);
A z = sq_dist(c, d);
A t = sq_dist(d, a);
return ((x == y) && (y == z) && (z == t));
}
template <typename A>
bool is_rectangle(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c,
const pair<A, A>& d) {
pair<A, A> x = a - b, y = b - c, z = c - d, t = d - a;
return ((x * y == 0) && (y * z == 0) && (z * t == 0) && (t * x == 0));
}
template <typename A>
bool are_parallel(const pair<A, A>& a, const pair<A, A>& b) {
return (a % b == 0);
}
template <typename A>
bool are_orthogonal(const pair<A, A>& a, const pair<A, A>& b) {
return (a * b == 0);
}
template <typename A>
bool are_codirected(const pair<A, A>& a, const pair<A, A>& b) {
return (are_parallel(a, b) && (a * b >= 0));
}
template <typename A>
bool is_parallelogram(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
return ((a - b) == (d - c));
}
template <typename A>
bool is_trapezoid(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c,
const pair<A, A>& d) {
pair<A, A> x = a - b, z = d - c;
return are_codirected(x, z);
}
template <typename A>
bool is_convex_polygon(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
pair<A, A> x = a - b, y = b - c, z = c - d, t = d - a;
A p = x % y, q = y % z, r = z % t, s = t % x;
return (((p >= 0) && (q >= 0) && (r >= 0) && (s >= 0)) ||
((p <= 0) && (q <= 0) && (r <= 0) && (s <= 0)));
}
template <typename A>
bool nprcs(const pair<A, A>& a, const pair<A, A>& c) {
return ((a % c) >= 0);
}
template <typename A>
bool nprcs(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return nprcs(a - b, a - c);
}
template <typename A>
bool npocs(const pair<A, A>& a, const pair<A, A>& c) {
return ((a % c) <= 0);
}
template <typename A>
bool npocs(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return npocs(a - b, a - c);
}
template <typename A>
bool prcs(const pair<A, A>& a, const pair<A, A>& c) {
return ((a % c) > 0);
}
template <typename A>
bool prcs(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return prcs(a - b, a - c);
}
template <typename A>
bool pocs(const pair<A, A>& a, const pair<A, A>& c) {
return ((a % c) < 0);
}
template <typename A>
bool pocs(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return pocs(a - b, a - c);
}
template <typename A>
bool different_sides(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
pair<A, A> x = b - a;
A p = x % (c - b), q = x % (d - b);
return (((p >= 0) && (q <= 0)) || ((p <= 0) && (q >= 0)));
}
template <typename A>
bool sharply_different_sides(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
pair<A, A> x = b - a;
A p = x % (c - b), q = x % (d - b);
return (((p > 0) && (q < 0)) || ((p < 0) && (q > 0)));
}
template <typename A>
pair<A, A> rot_90(const pair<A, A>& p) {
return make_pair(-p.second, p.first);
}
template <typename A>
pair<A, A> rot_90(const pair<A, A>& p, const pair<A, A>& c) {
return c + rot_90(p - c);
}
template <typename A>
bool intersecting_segments(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
return different_sides(a, b, c, d) && different_sides(c, d, a, b);
}
template <typename A>
bool sharply_intersecting_segments(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
return sharply_different_sides(a, b, c, d) &&
sharply_different_sides(c, d, a, b);
}
template <typename A>
pair<pair<A, A>, A> line_with_normal_vector_through_point(
const pair<A, A>& normal, const pair<A, A>& point) {
return make_pair(normal, -(normal * point));
}
template <typename A>
pair<pair<A, A>, A> serper(const pair<A, A>& a, const pair<A, A>& b) {
pair<A, A> p = b - a;
return make_pair(p + p, -(p * (a + b)));
}
pair<double, double> ints(const pair<double, double>& p, const double& a,
const pair<double, double>& q, const double& b) {
double D = p % q;
double E =
pair<double, double>(-a, p.second) % pair<double, double>(-b, q.second);
double F =
pair<double, double>(p.first, -a) % pair<double, double>(q.first, -b);
return pair<double, double>(E / D, F / D);
}
pair<double, double> circumcenter(const pair<double, double>& x,
const pair<double, double>& y,
const pair<double, double>& z) {
auto p1 = serper(x, y), p2 = serper(x, z);
return ints(p1.first, p1.second, p2.first, p2.second);
}
template <typename A>
pair<pair<A, A>, A> l_th_2(const pair<A, A>& p, const pair<A, A>& q) {
return make_pair(make_pair(q.second - p.second, p.first - q.first), -p % q);
}
template <typename A>
vector<pair<double, double>> circle_intersection(const pair<pair<A, A>, A>& a,
const pair<pair<A, A>, A>& b) {
pair<A, A> c = b.first - a.first;
A rr1 = a.second * a.second, rr2 = b.second * b.second, cc = c * c,
rrrr1 = rr1 * rr1, rrrr2 = rr2 * rr2, cccc = cc * cc,
D = 2 * (rr1 * rr2 + rr2 * cc + cc * rr1) - (rrrr1 + rrrr2 + cccc);
if (D >= 0) {
double E = (((double)(rr1 - rr2)) / cc + 1) / 2;
pair<double, double> baza =
pair<double, double>(a.first.first, a.first.second) +
pair<double, double>(c.first, c.second) * E;
if (D) {
double lll = sqrt((double)(D)) / (2 * cc);
pair<A, A> cr = rot_90(c);
pair<double, double> pmm =
pair<double, double>(cr.first, cr.second) * lll;
return {baza + pmm, baza - pmm};
} else
return vector<pair<double, double>>(1, baza);
} else
return vector<pair<double, double>>();
}
template <typename A, typename B>
pair<A, B> sopr(const pair<A, B>& p) {
return make_pair(p.first, -p.second);
}
template <typename A>
bool on_segment(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return are_codirected(b - a, c - b);
}
template <typename A>
pair<pair<A, A>, A> perpendicular(const pair<A, A>& line,
const pair<A, A>& point) {
return line_with_normal_vector_through_point(rot_90(line), point);
}
pair<double, double> projection(const pair<pair<double, double>, double>& line,
const pair<double, double>& point) {
pair<pair<double, double>, double> temp = perpendicular(line.first, point);
return ints(line.first, line.second, temp.first, temp.second);
}
pair<double, double> height(const pair<double, double>& a,
const pair<double, double>& b,
const pair<double, double>& c) {
return projection(l_th_2(a, c), b);
}
pair<double, double> unitvector(const pair<double, double>& a) {
return a * (1 / len(a));
}
template <typename T>
vector<int> z_function(const vector<T>& s) {
int n = ((int)(s.size()));
vector<int> z(n);
int l = 0, r = 1;
for (int i = 1; i < n; ++i) {
z[i] = max(0, min(r - i, z[i - l]));
while (i + z[i] < n && (s[i + z[i]] == s[z[i]])) ++z[i];
if (r < i + z[i]) {
l = i;
r = i + z[i];
}
}
return z;
}
pair<int, int> euc(const int& m, const int& n) {
if (n == 0) return pair<int, int>((m >= 0) ? 1 : -1, 0);
int t = m / n;
pair<int, int> ans1 = euc(n, m - t * n);
return pair<int, int>(ans1.second, ans1.first - ans1.second * t);
}
int prod(const int& a, const int& b, const int& M) {
return ((long long)(a)) * b % M;
}
int sum(const int& a, const int& b, const int& M) {
int c = a + b;
return c >= M ? c - M : c;
}
int raz(const int& a, const int& b, const int& M) {
int c = a - b;
return c < 0 ? c + M : c;
}
long long prodll(const long long& a, const long long& b, const long long& M) {
return a * b % M;
}
long long sumll(const long long& a, const long long& b, const long long& M) {
long long c = a + b;
return c >= M ? c - M : c;
}
long long razll(const long long& a, const long long& b, const long long& M) {
long long c = a - b;
return c < 0 ? c + M : c;
}
template <typename A>
bool angdis_cmp(const pair<A, A>& a, const pair<A, A>& b) {
A p;
if (p = a % b)
return (p > 0);
else
return sq_len(a) < sq_len(b);
}
template <typename T>
int find_min_idx(const vector<T>& v) {
int ans = 0, n = ((int)(v.size()));
for (int i = 1; i < n; ++i)
if (v[i] < v[ans]) ans = i;
return ans;
}
template <typename T>
vector<int> convex_hull(vector<pair<T, T>>& a) {
int n = ((int)(a.size()));
if (n) {
int mt = find_min_idx(a);
pair<T, T> d = a[mt];
for (int i = 0; i < n; ++i) a[i] -= d;
vector<int> idx(n);
for (int i = 0; i < n; ++i) idx[i] = i;
sort(idx.begin(), idx.end(),
[&](const int& l, const int& r) { return angdis_cmp(a[l], a[r]); });
vector<int> h(1, idx.front());
for (auto it = idx.begin() + 1; it != idx.end(); ++it) {
auto temp = a.begin() + *it;
if (((int)(h.size())) >= 2) {
if (are_parallel(a[h.back()], *temp)) h.pop_back();
while ((((int)(h.size())) >= 3) &&
npocs(a[h[((int)(h.size())) - 2]], a[h.back()], *temp))
h.pop_back();
}
h.push_back(*it);
}
for (int i = 0; i < n; ++i) a[i] += d;
return h;
} else
return vector<int>();
}
pair<int, int> cool_gcd(const int& a, const int& b) {
if (b) {
int c = a / b;
pair<int, int> ans1 = cool_gcd(b, a - b * c);
return pair<long long, long long>(ans1.second,
ans1.first - ans1.second * c);
} else
return pair<int, int>(1, 0);
}
pair<long long, long long> cool_gcdll(const long long& a, const long long& b) {
if (b) {
long long c = a / b;
pair<long long, long long> ans1 = cool_gcdll(b, a - b * c);
return pair<long long, long long>(ans1.second,
ans1.first - ans1.second * c);
} else
return pair<long long, long long>(1ll, 0ll);
}
int gcd(int a, int b) {
while (b) {
a %= b;
swap(a, b);
}
return a;
}
long long gcdll(long long a, long long b) {
while (b) {
a %= b;
swap(a, b);
}
return a;
}
long long pr_p(const long long& a, const long long& b, const long long& p) {
if (b < 1000000) return (a * b) % p;
long long temp = pr_p(a, b >> 1ll, p);
temp = sumll(temp, temp, p);
if (b & 1ll)
return sumll(temp, a, p);
else
return temp;
}
long long po_p(const long long& a, const long long& b, const long long& p) {
if (b < 2) {
if (b == 0)
return 1;
else
return a;
}
long long temp = po_p(a, b >> 1ll, p);
temp = pr_p(temp, temp, p);
if (b & 1ll)
return pr_p(temp, a, p);
else
return temp;
}
int pow_mod(const int& a, const int& b, const int& p) {
if (b < 2) {
if (b == 0)
return 1;
else
return a;
}
int temp = pow_mod(a, b >> 1, p);
temp = prod(temp, temp, p);
if (b & 1)
return prod(temp, a, p);
else
return temp;
}
long long pow_modll(const long long& a, const long long& b,
const long long& p) {
if (b < 2) {
if (b == 0)
return 1;
else
return a;
}
long long temp = pow_modll(a, b >> 1ll, p);
temp = prodll(temp, temp, p);
if (b & 1ll)
return prodll(temp, a, p);
else
return temp;
}
int inverse(int a, int n) {
int c = cool_gcd(a, n).first;
if (c < 0) c += n;
return c;
}
long long inversell(long long a, long long n) {
long long c = cool_gcdll(a, n).first;
if (c < 0) c += n;
return c;
}
template <typename A>
pair<int, int> equal_elements(const vector<A>& u, const vector<A>& v) {
pair<int, int> ans(INT_MAX, INT_MAX);
int m = ((int)(u.size())), n = ((int)(v.size()));
vector<int> id_u(m);
for (int i = 1; i < m; ++i) id_u[i] = i;
vector<int> id_v(n);
for (int i = 1; i < n; ++i) id_v[i] = i;
sort(id_u.begin(), id_u.end(),
[&](const int& x, const int& y) { return u[x] < u[y]; });
sort(id_v.begin(), id_v.end(),
[&](const int& x, const int& y) { return v[x] < v[y]; });
int i = 0;
int j = 0;
while ((i < m) && (j < n)) {
if (u[id_u[i]] < v[id_v[j]])
++i;
else if (v[id_v[j]] < u[id_u[i]])
++j;
else {
ans = min(ans, pair<int, int>(id_v[j], id_u[i]));
++j;
}
}
if (ans.first == INT_MAX)
return pair<int, int>(-1, -1);
else
return pair<int, int>(ans.second, ans.first);
}
long long discr_log(long long a, long long b, long long n) {
int k = ((int)(sqrt((long double)n)));
long long a1 = inversell(a, n);
int l = k + 20;
long long a2 = po_p(a1, k, n);
vector<long long> seq1(k), seq2(l);
seq1.front() = 1;
for (int i = 1; i < k; ++i) seq1[i] = pr_p(seq1[i - 1], a, n);
seq2.front() = b;
for (int i = 1; i < l; ++i) seq2[i] = pr_p(seq2[i - 1], a2, n);
pair<long long, long long> e = equal_elements(seq1, seq2);
if (e.first == -1)
return -1;
else
return e.first + e.second * k;
}
long long common_discr_log(long long a, long long b, long long n) {
const int C = 70;
a %= n;
b %= n;
if (gcdll(n, a) != 1) {
for (int i = 0; i < C; ++i) {
if (po_p(a, i, n) == b) {
return i;
}
}
long long multp = po_p(a, C, n);
long long g = gcdll(multp, n);
if (b % g) {
return -1;
} else {
b /= g;
n /= g;
multp /= g;
long long pop_back = inversell(multp, n);
b = pr_p(b, pop_back, n);
long long ans = discr_log(a, b, n);
if (ans == -1)
return -1;
else
return ans + C;
}
} else
return discr_log(a, b, n);
}
struct factorizator {
const int N = 2000001;
vector<int> pr;
vector<int> md;
vector<int> pw;
vector<int> without_md;
void init() {
md.resize(N);
pw.resize(N);
pr.reserve(N);
without_md.resize(N, 1);
for (int i = 2; i < N; ++i) md[i] = i;
for (int i = 2; i < N; ++i) {
if (md[i] == i) pr.push_back(i);
bool worth = true;
for (int j = 0; worth && (j < ((int)(pr.size()))) && (pr[j] <= md[i]);
++j) {
long long temp = ((long long)(pr[j])) * i;
if (temp < N)
md[((int)temp)] = pr[j];
else
worth = false;
}
}
for (int i = 2; i < N; ++i) {
int t = md[i], s = i / t;
if (md[s] == t) {
pw[i] = 1 + pw[s];
without_md[i] = without_md[s];
} else {
pw[i] = 1;
without_md[i] = s;
}
}
}
void known_factorization(int n, vector<pair<int, int>>& v) {
while (n > 1) {
v.emplace_back(md[n], pw[n]);
n = without_md[n];
}
}
vector<pair<int, int>> factorize(int n) {
vector<pair<int, int>> ans;
for (int i = 0;
(i < ((int)(pr.size()))) && (pr[i] * ((long long)(pr[i])) <= n); ++i) {
int t = n / pr[i];
if (t * pr[i] == n) {
ans.emplace_back(pr[i], 0);
do {
++ans.back().second;
n = t;
t = n / pr[i];
} while (t * pr[i] == n);
}
if (n < N) {
known_factorization(n, ans);
return ans;
}
}
if (n > 1) ans.emplace_back(n, 1);
return ans;
}
vector<pair<int, int>> merge(const vector<pair<int, int>>& p1,
const vector<pair<int, int>>& p2) {
vector<pair<int, int>> ans;
int m = ((int)(p1.size()));
int n = ((int)(p2.size()));
int i = 0;
int j = 0;
while ((i < m) || (j < n)) {
if (i < m) {
if (j < n) {
if (p1[i].first < p2[j].first)
ans.push_back(p1[i++]);
else if (p1[i].first > p2[j].first)
ans.push_back(p2[j++]);
else {
ans.emplace_back(p1[i].first, p1[i].second + p2[j].second);
++i;
++j;
}
} else
ans.push_back(p1[i++]);
} else
ans.push_back(p2[j++]);
}
}
};
factorizator fac;
int phi(const vector<pair<int, int>>& v) {
int ans = 1;
for (int i = 0; i < ((int)(v.size())); ++i) {
ans *= v[i].first - 1;
for (int j = 1; j < v[i].second; ++j) ans *= v[i].first;
}
return ans;
}
int phi(const int& n) { return phi(fac.factorize(n)); }
bool check_primitive_root(const int& ph, const vector<int>& to_check,
const int& r, const int& n) {
for (int i = 0; i < ((int)(to_check.size())); ++i)
if (pow_mod(r, to_check[i], n) == 1) return false;
return (pow_mod(r, ph, n) == 1);
}
int primitive_root(const int& n) {
if (n < 3) return n - 1;
int p = phi(n);
vector<pair<int, int>> f = fac.factorize(p);
vector<int> to_check(((int)(f.size())));
for (int i = 0; i < ((int)(f.size())); ++i) to_check[i] = p / f[i].first;
for (int i = 2; i < n; ++i)
if (check_primitive_root(p, to_check, i, n)) return i;
return -1;
}
int unite_mod(const int& a, const int& p, const int& b, const int& q) {
pair<int, int> c = cool_gcd(p, q);
int pr = p * q;
int ans = ((a * c.second * q + b * c.first * p) % pr + pr) % pr;
return ans;
}
long long unite_modll(const long long& a, const long long& p,
const long long& b, const long long& q) {
pair<long long, long long> c = cool_gcdll(p, q);
long long pr = p * q;
long long ans = ((a * c.second * q + b * c.first * p) % pr + pr) % pr;
return ans;
}
pair<int, int> power_v(int n, const int& p) {
int ans = 0;
while (n % p == 0) {
n /= p;
++ans;
}
return pair<int, int>(ans, n);
}
int square_root_prime_modulo(int c, int n, const int& pr, const int& k) {
c %= n;
if (c) {
pair<int, int> kek = power_v(c, pr);
int l = kek.first;
if (l & 1) return -1;
if (l > 0) {
int pwl = 1;
for (int i = 0; i < l; ++i) pwl *= pr;
n /= pwl;
c /= pwl;
int ans1 = square_root_prime_modulo(c, n, pr, k - l);
if (ans1 == -1) return -1;
for (int i = 0; i < (l >> 1); ++i) ans1 *= pr;
return ans1;
} else {
int primitive;
if (n & 1)
primitive = primitive_root(n);
else
primitive = 5;
int u = ((int)discr_log(primitive, c, n));
if (u == -1) return -1;
if (u & 1) return -1;
return pow_mod(primitive, u >> 1, n);
}
} else
return 0;
}
int square_root_modulo(const int& c, const int& n) {
vector<pair<int, int>> f = fac.factorize(n);
int a = 0, p = 1;
for (int i = 0; i < ((int)(f.size())); ++i) {
int q = 1;
for (int j = 0; j < f[i].second; ++j) q *= f[i].first;
int b = square_root_prime_modulo(c, q, f[i].first, f[i].second);
if (b == -1) return -1;
a = unite_mod(a, p, b, q);
p *= q;
}
return a;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cerr.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2) && (y1 <= y2))
cout << "Polycarp\n";
else if (x1 + y1 <= max(x2, y2))
cout << "Polycarp\n";
else
cout << "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>
#pragma gcc optimize("O3")
#pragma gcc target("sse4")
using namespace std;
mt19937 rng(2391);
template <typename A>
istream& operator>>(istream& fin, vector<A>& v) {
for (auto it = v.begin(); it != v.end(); ++it) fin >> *it;
return fin;
}
template <typename A, typename B>
istream& operator>>(istream& fin, pair<A, B>& p) {
fin >> p.first >> p.second;
return fin;
}
template <typename A, typename B>
pair<A, B> operator+(const pair<A, B>& a, const pair<A, B>& b) {
return make_pair(a.first + b.first, a.second + b.second);
}
template <typename A, typename B>
pair<A, B> operator+=(pair<A, B>& a, const pair<A, B>& b) {
a.first += b.first;
a.second += b.second;
return a;
}
template <typename A, typename B>
pair<A, B> operator-(const pair<A, B>& a, const pair<A, B>& b) {
return make_pair(a.first - b.first, a.second - b.second);
}
template <typename A, typename B>
pair<A, B> operator-(const pair<A, B>& a) {
return make_pair(-a.first, -a.second);
}
template <typename A, typename B>
pair<A, B>& operator++(pair<A, B>& a) {
++a.first;
++a.second;
return a;
}
template <typename A, typename B>
pair<A, B>& operator--(pair<A, B>& a) {
--a.first;
--a.second;
return a;
}
template <typename A>
vector<A>& operator++(vector<A>& a) {
for (auto it = a.begin(); it != a.end(); ++it) ++*it;
return a;
}
template <typename A>
vector<A>& operator--(vector<A>& a) {
for (auto it = a.begin(); it != a.end(); ++it) --*it;
return a;
}
template <typename A, typename B>
pair<A, B> operator++(pair<A, B>& a, int) {
pair<A, B> b = a;
++a;
return b;
}
template <typename A, typename B>
pair<A, B> operator--(pair<A, B>& a, int) {
pair<A, B> b = a;
--a;
return b;
}
template <typename A>
vector<A>& operator++(vector<A>& a, int) {
vector<A> b = a;
++a;
return b;
}
template <typename A>
vector<A>& operator--(vector<A>& a, int) {
vector<A> b = a;
--a;
return b;
}
vector<vector<int>> adjlist_from_edgelist(const vector<pair<int, int>>& e,
const int& n) {
vector<vector<int>> g(n);
for (auto it = e.begin(); it != e.end(); ++it) {
g[it->first].push_back(it->second);
g[it->second].push_back(it->first);
}
return g;
}
template <typename A, typename B>
pair<A, B> operator-=(pair<A, B>& a, const pair<A, B>& b) {
a.first -= b.first;
a.second -= b.second;
return a;
}
template <typename A>
A operator*(const pair<A, A>& p, const pair<A, A>& q) {
return p.first * q.first + p.second * q.second;
}
template <typename A>
pair<A, A> operator*(const pair<A, A>& p, const A& q) {
return make_pair(p.first * q, p.second * q);
}
template <typename A>
A operator%(const pair<A, A>& p, const pair<A, A>& q) {
return p.first * q.second - p.second * q.first;
}
template <typename A>
A sq_len(const pair<A, A>& p) {
return p * p;
}
template <typename A>
A sq_dist(const pair<A, A>& p, const pair<A, A>& q) {
return sq_len(p - q);
}
template <typename A>
double len(const pair<A, A>& p) {
return sqrt(sq_len(p));
}
template <typename A>
double dist(const pair<A, A>& p, const pair<A, A>& q) {
return len(p - q);
}
template <typename A>
bool is_rhombus(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c,
const pair<A, A>& d) {
A x = sq_dist(a, b);
A y = sq_dist(b, c);
A z = sq_dist(c, d);
A t = sq_dist(d, a);
return ((x == y) && (y == z) && (z == t));
}
template <typename A>
bool is_rectangle(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c,
const pair<A, A>& d) {
pair<A, A> x = a - b, y = b - c, z = c - d, t = d - a;
return ((x * y == 0) && (y * z == 0) && (z * t == 0) && (t * x == 0));
}
template <typename A>
bool are_parallel(const pair<A, A>& a, const pair<A, A>& b) {
return (a % b == 0);
}
template <typename A>
bool are_orthogonal(const pair<A, A>& a, const pair<A, A>& b) {
return (a * b == 0);
}
template <typename A>
bool are_codirected(const pair<A, A>& a, const pair<A, A>& b) {
return (are_parallel(a, b) && (a * b >= 0));
}
template <typename A>
bool is_parallelogram(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
return ((a - b) == (d - c));
}
template <typename A>
bool is_trapezoid(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c,
const pair<A, A>& d) {
pair<A, A> x = a - b, z = d - c;
return are_codirected(x, z);
}
template <typename A>
bool is_convex_polygon(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
pair<A, A> x = a - b, y = b - c, z = c - d, t = d - a;
A p = x % y, q = y % z, r = z % t, s = t % x;
return (((p >= 0) && (q >= 0) && (r >= 0) && (s >= 0)) ||
((p <= 0) && (q <= 0) && (r <= 0) && (s <= 0)));
}
template <typename A>
bool nprcs(const pair<A, A>& a, const pair<A, A>& c) {
return ((a % c) >= 0);
}
template <typename A>
bool nprcs(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return nprcs(a - b, a - c);
}
template <typename A>
bool npocs(const pair<A, A>& a, const pair<A, A>& c) {
return ((a % c) <= 0);
}
template <typename A>
bool npocs(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return npocs(a - b, a - c);
}
template <typename A>
bool prcs(const pair<A, A>& a, const pair<A, A>& c) {
return ((a % c) > 0);
}
template <typename A>
bool prcs(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return prcs(a - b, a - c);
}
template <typename A>
bool pocs(const pair<A, A>& a, const pair<A, A>& c) {
return ((a % c) < 0);
}
template <typename A>
bool pocs(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return pocs(a - b, a - c);
}
template <typename A>
bool different_sides(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
pair<A, A> x = b - a;
A p = x % (c - b), q = x % (d - b);
return (((p >= 0) && (q <= 0)) || ((p <= 0) && (q >= 0)));
}
template <typename A>
bool sharply_different_sides(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
pair<A, A> x = b - a;
A p = x % (c - b), q = x % (d - b);
return (((p > 0) && (q < 0)) || ((p < 0) && (q > 0)));
}
template <typename A>
pair<A, A> rot_90(const pair<A, A>& p) {
return make_pair(-p.second, p.first);
}
template <typename A>
pair<A, A> rot_90(const pair<A, A>& p, const pair<A, A>& c) {
return c + rot_90(p - c);
}
template <typename A>
bool intersecting_segments(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
return different_sides(a, b, c, d) && different_sides(c, d, a, b);
}
template <typename A>
bool sharply_intersecting_segments(const pair<A, A>& a, const pair<A, A>& b,
const pair<A, A>& c, const pair<A, A>& d) {
return sharply_different_sides(a, b, c, d) &&
sharply_different_sides(c, d, a, b);
}
template <typename A>
pair<pair<A, A>, A> line_with_normal_vector_through_point(
const pair<A, A>& normal, const pair<A, A>& point) {
return make_pair(normal, -(normal * point));
}
template <typename A>
pair<pair<A, A>, A> serper(const pair<A, A>& a, const pair<A, A>& b) {
pair<A, A> p = b - a;
return make_pair(p + p, -(p * (a + b)));
}
pair<double, double> ints(const pair<double, double>& p, const double& a,
const pair<double, double>& q, const double& b) {
double D = p % q;
double E =
pair<double, double>(-a, p.second) % pair<double, double>(-b, q.second);
double F =
pair<double, double>(p.first, -a) % pair<double, double>(q.first, -b);
return pair<double, double>(E / D, F / D);
}
pair<double, double> circumcenter(const pair<double, double>& x,
const pair<double, double>& y,
const pair<double, double>& z) {
auto p1 = serper(x, y), p2 = serper(x, z);
return ints(p1.first, p1.second, p2.first, p2.second);
}
template <typename A>
pair<pair<A, A>, A> l_th_2(const pair<A, A>& p, const pair<A, A>& q) {
return make_pair(make_pair(q.second - p.second, p.first - q.first), -p % q);
}
template <typename A>
vector<pair<double, double>> circle_intersection(const pair<pair<A, A>, A>& a,
const pair<pair<A, A>, A>& b) {
pair<A, A> c = b.first - a.first;
A rr1 = a.second * a.second, rr2 = b.second * b.second, cc = c * c,
rrrr1 = rr1 * rr1, rrrr2 = rr2 * rr2, cccc = cc * cc,
D = 2 * (rr1 * rr2 + rr2 * cc + cc * rr1) - (rrrr1 + rrrr2 + cccc);
if (D >= 0) {
double E = (((double)(rr1 - rr2)) / cc + 1) / 2;
pair<double, double> baza =
pair<double, double>(a.first.first, a.first.second) +
pair<double, double>(c.first, c.second) * E;
if (D) {
double lll = sqrt((double)(D)) / (2 * cc);
pair<A, A> cr = rot_90(c);
pair<double, double> pmm =
pair<double, double>(cr.first, cr.second) * lll;
return {baza + pmm, baza - pmm};
} else
return vector<pair<double, double>>(1, baza);
} else
return vector<pair<double, double>>();
}
template <typename A, typename B>
pair<A, B> sopr(const pair<A, B>& p) {
return make_pair(p.first, -p.second);
}
template <typename A>
bool on_segment(const pair<A, A>& a, const pair<A, A>& b, const pair<A, A>& c) {
return are_codirected(b - a, c - b);
}
template <typename A>
pair<pair<A, A>, A> perpendicular(const pair<A, A>& line,
const pair<A, A>& point) {
return line_with_normal_vector_through_point(rot_90(line), point);
}
pair<double, double> projection(const pair<pair<double, double>, double>& line,
const pair<double, double>& point) {
pair<pair<double, double>, double> temp = perpendicular(line.first, point);
return ints(line.first, line.second, temp.first, temp.second);
}
pair<double, double> height(const pair<double, double>& a,
const pair<double, double>& b,
const pair<double, double>& c) {
return projection(l_th_2(a, c), b);
}
pair<double, double> unitvector(const pair<double, double>& a) {
return a * (1 / len(a));
}
template <typename T>
vector<int> z_function(const vector<T>& s) {
int n = ((int)(s.size()));
vector<int> z(n);
int l = 0, r = 1;
for (int i = 1; i < n; ++i) {
z[i] = max(0, min(r - i, z[i - l]));
while (i + z[i] < n && (s[i + z[i]] == s[z[i]])) ++z[i];
if (r < i + z[i]) {
l = i;
r = i + z[i];
}
}
return z;
}
pair<int, int> euc(const int& m, const int& n) {
if (n == 0) return pair<int, int>((m >= 0) ? 1 : -1, 0);
int t = m / n;
pair<int, int> ans1 = euc(n, m - t * n);
return pair<int, int>(ans1.second, ans1.first - ans1.second * t);
}
int prod(const int& a, const int& b, const int& M) {
return ((long long)(a)) * b % M;
}
int sum(const int& a, const int& b, const int& M) {
int c = a + b;
return c >= M ? c - M : c;
}
int raz(const int& a, const int& b, const int& M) {
int c = a - b;
return c < 0 ? c + M : c;
}
long long prodll(const long long& a, const long long& b, const long long& M) {
return a * b % M;
}
long long sumll(const long long& a, const long long& b, const long long& M) {
long long c = a + b;
return c >= M ? c - M : c;
}
long long razll(const long long& a, const long long& b, const long long& M) {
long long c = a - b;
return c < 0 ? c + M : c;
}
template <typename A>
bool angdis_cmp(const pair<A, A>& a, const pair<A, A>& b) {
A p;
if (p = a % b)
return (p > 0);
else
return sq_len(a) < sq_len(b);
}
template <typename T>
int find_min_idx(const vector<T>& v) {
int ans = 0, n = ((int)(v.size()));
for (int i = 1; i < n; ++i)
if (v[i] < v[ans]) ans = i;
return ans;
}
template <typename T>
vector<int> convex_hull(vector<pair<T, T>>& a) {
int n = ((int)(a.size()));
if (n) {
int mt = find_min_idx(a);
pair<T, T> d = a[mt];
for (int i = 0; i < n; ++i) a[i] -= d;
vector<int> idx(n);
for (int i = 0; i < n; ++i) idx[i] = i;
sort(idx.begin(), idx.end(),
[&](const int& l, const int& r) { return angdis_cmp(a[l], a[r]); });
vector<int> h(1, idx.front());
for (auto it = idx.begin() + 1; it != idx.end(); ++it) {
auto temp = a.begin() + *it;
if (((int)(h.size())) >= 2) {
if (are_parallel(a[h.back()], *temp)) h.pop_back();
while ((((int)(h.size())) >= 3) &&
npocs(a[h[((int)(h.size())) - 2]], a[h.back()], *temp))
h.pop_back();
}
h.push_back(*it);
}
for (int i = 0; i < n; ++i) a[i] += d;
return h;
} else
return vector<int>();
}
pair<int, int> cool_gcd(const int& a, const int& b) {
if (b) {
int c = a / b;
pair<int, int> ans1 = cool_gcd(b, a - b * c);
return pair<long long, long long>(ans1.second,
ans1.first - ans1.second * c);
} else
return pair<int, int>(1, 0);
}
pair<long long, long long> cool_gcdll(const long long& a, const long long& b) {
if (b) {
long long c = a / b;
pair<long long, long long> ans1 = cool_gcdll(b, a - b * c);
return pair<long long, long long>(ans1.second,
ans1.first - ans1.second * c);
} else
return pair<long long, long long>(1ll, 0ll);
}
int gcd(int a, int b) {
while (b) {
a %= b;
swap(a, b);
}
return a;
}
long long gcdll(long long a, long long b) {
while (b) {
a %= b;
swap(a, b);
}
return a;
}
long long pr_p(const long long& a, const long long& b, const long long& p) {
if (b < 1000000) return (a * b) % p;
long long temp = pr_p(a, b >> 1ll, p);
temp = sumll(temp, temp, p);
if (b & 1ll)
return sumll(temp, a, p);
else
return temp;
}
long long po_p(const long long& a, const long long& b, const long long& p) {
if (b < 2) {
if (b == 0)
return 1;
else
return a;
}
long long temp = po_p(a, b >> 1ll, p);
temp = pr_p(temp, temp, p);
if (b & 1ll)
return pr_p(temp, a, p);
else
return temp;
}
int pow_mod(const int& a, const int& b, const int& p) {
if (b < 2) {
if (b == 0)
return 1;
else
return a;
}
int temp = pow_mod(a, b >> 1, p);
temp = prod(temp, temp, p);
if (b & 1)
return prod(temp, a, p);
else
return temp;
}
long long pow_modll(const long long& a, const long long& b,
const long long& p) {
if (b < 2) {
if (b == 0)
return 1;
else
return a;
}
long long temp = pow_modll(a, b >> 1ll, p);
temp = prodll(temp, temp, p);
if (b & 1ll)
return prodll(temp, a, p);
else
return temp;
}
int inverse(int a, int n) {
int c = cool_gcd(a, n).first;
if (c < 0) c += n;
return c;
}
long long inversell(long long a, long long n) {
long long c = cool_gcdll(a, n).first;
if (c < 0) c += n;
return c;
}
template <typename A>
pair<int, int> equal_elements(const vector<A>& u, const vector<A>& v) {
pair<int, int> ans(INT_MAX, INT_MAX);
int m = ((int)(u.size())), n = ((int)(v.size()));
vector<int> id_u(m);
for (int i = 1; i < m; ++i) id_u[i] = i;
vector<int> id_v(n);
for (int i = 1; i < n; ++i) id_v[i] = i;
sort(id_u.begin(), id_u.end(),
[&](const int& x, const int& y) { return u[x] < u[y]; });
sort(id_v.begin(), id_v.end(),
[&](const int& x, const int& y) { return v[x] < v[y]; });
int i = 0;
int j = 0;
while ((i < m) && (j < n)) {
if (u[id_u[i]] < v[id_v[j]])
++i;
else if (v[id_v[j]] < u[id_u[i]])
++j;
else {
ans = min(ans, pair<int, int>(id_v[j], id_u[i]));
++j;
}
}
if (ans.first == INT_MAX)
return pair<int, int>(-1, -1);
else
return pair<int, int>(ans.second, ans.first);
}
long long discr_log(long long a, long long b, long long n) {
int k = ((int)(sqrt((long double)n)));
long long a1 = inversell(a, n);
int l = k + 20;
long long a2 = po_p(a1, k, n);
vector<long long> seq1(k), seq2(l);
seq1.front() = 1;
for (int i = 1; i < k; ++i) seq1[i] = pr_p(seq1[i - 1], a, n);
seq2.front() = b;
for (int i = 1; i < l; ++i) seq2[i] = pr_p(seq2[i - 1], a2, n);
pair<long long, long long> e = equal_elements(seq1, seq2);
if (e.first == -1)
return -1;
else
return e.first + e.second * k;
}
long long common_discr_log(long long a, long long b, long long n) {
const int C = 70;
a %= n;
b %= n;
if (gcdll(n, a) != 1) {
for (int i = 0; i < C; ++i) {
if (po_p(a, i, n) == b) {
return i;
}
}
long long multp = po_p(a, C, n);
long long g = gcdll(multp, n);
if (b % g) {
return -1;
} else {
b /= g;
n /= g;
multp /= g;
long long pop_back = inversell(multp, n);
b = pr_p(b, pop_back, n);
long long ans = discr_log(a, b, n);
if (ans == -1)
return -1;
else
return ans + C;
}
} else
return discr_log(a, b, n);
}
struct factorizator {
const int N = 2000001;
vector<int> pr;
vector<int> md;
vector<int> pw;
vector<int> without_md;
void init() {
md.resize(N);
pw.resize(N);
pr.reserve(N);
without_md.resize(N, 1);
for (int i = 2; i < N; ++i) md[i] = i;
for (int i = 2; i < N; ++i) {
if (md[i] == i) pr.push_back(i);
bool worth = true;
for (int j = 0; worth && (j < ((int)(pr.size()))) && (pr[j] <= md[i]);
++j) {
long long temp = ((long long)(pr[j])) * i;
if (temp < N)
md[((int)temp)] = pr[j];
else
worth = false;
}
}
for (int i = 2; i < N; ++i) {
int t = md[i], s = i / t;
if (md[s] == t) {
pw[i] = 1 + pw[s];
without_md[i] = without_md[s];
} else {
pw[i] = 1;
without_md[i] = s;
}
}
}
void known_factorization(int n, vector<pair<int, int>>& v) {
while (n > 1) {
v.emplace_back(md[n], pw[n]);
n = without_md[n];
}
}
vector<pair<int, int>> factorize(int n) {
vector<pair<int, int>> ans;
for (int i = 0;
(i < ((int)(pr.size()))) && (pr[i] * ((long long)(pr[i])) <= n); ++i) {
int t = n / pr[i];
if (t * pr[i] == n) {
ans.emplace_back(pr[i], 0);
do {
++ans.back().second;
n = t;
t = n / pr[i];
} while (t * pr[i] == n);
}
if (n < N) {
known_factorization(n, ans);
return ans;
}
}
if (n > 1) ans.emplace_back(n, 1);
return ans;
}
vector<pair<int, int>> merge(const vector<pair<int, int>>& p1,
const vector<pair<int, int>>& p2) {
vector<pair<int, int>> ans;
int m = ((int)(p1.size()));
int n = ((int)(p2.size()));
int i = 0;
int j = 0;
while ((i < m) || (j < n)) {
if (i < m) {
if (j < n) {
if (p1[i].first < p2[j].first)
ans.push_back(p1[i++]);
else if (p1[i].first > p2[j].first)
ans.push_back(p2[j++]);
else {
ans.emplace_back(p1[i].first, p1[i].second + p2[j].second);
++i;
++j;
}
} else
ans.push_back(p1[i++]);
} else
ans.push_back(p2[j++]);
}
}
};
factorizator fac;
int phi(const vector<pair<int, int>>& v) {
int ans = 1;
for (int i = 0; i < ((int)(v.size())); ++i) {
ans *= v[i].first - 1;
for (int j = 1; j < v[i].second; ++j) ans *= v[i].first;
}
return ans;
}
int phi(const int& n) { return phi(fac.factorize(n)); }
bool check_primitive_root(const int& ph, const vector<int>& to_check,
const int& r, const int& n) {
for (int i = 0; i < ((int)(to_check.size())); ++i)
if (pow_mod(r, to_check[i], n) == 1) return false;
return (pow_mod(r, ph, n) == 1);
}
int primitive_root(const int& n) {
if (n < 3) return n - 1;
int p = phi(n);
vector<pair<int, int>> f = fac.factorize(p);
vector<int> to_check(((int)(f.size())));
for (int i = 0; i < ((int)(f.size())); ++i) to_check[i] = p / f[i].first;
for (int i = 2; i < n; ++i)
if (check_primitive_root(p, to_check, i, n)) return i;
return -1;
}
int unite_mod(const int& a, const int& p, const int& b, const int& q) {
pair<int, int> c = cool_gcd(p, q);
int pr = p * q;
int ans = ((a * c.second * q + b * c.first * p) % pr + pr) % pr;
return ans;
}
long long unite_modll(const long long& a, const long long& p,
const long long& b, const long long& q) {
pair<long long, long long> c = cool_gcdll(p, q);
long long pr = p * q;
long long ans = ((a * c.second * q + b * c.first * p) % pr + pr) % pr;
return ans;
}
pair<int, int> power_v(int n, const int& p) {
int ans = 0;
while (n % p == 0) {
n /= p;
++ans;
}
return pair<int, int>(ans, n);
}
int square_root_prime_modulo(int c, int n, const int& pr, const int& k) {
c %= n;
if (c) {
pair<int, int> kek = power_v(c, pr);
int l = kek.first;
if (l & 1) return -1;
if (l > 0) {
int pwl = 1;
for (int i = 0; i < l; ++i) pwl *= pr;
n /= pwl;
c /= pwl;
int ans1 = square_root_prime_modulo(c, n, pr, k - l);
if (ans1 == -1) return -1;
for (int i = 0; i < (l >> 1); ++i) ans1 *= pr;
return ans1;
} else {
int primitive;
if (n & 1)
primitive = primitive_root(n);
else
primitive = 5;
int u = ((int)discr_log(primitive, c, n));
if (u == -1) return -1;
if (u & 1) return -1;
return pow_mod(primitive, u >> 1, n);
}
} else
return 0;
}
int square_root_modulo(const int& c, const int& n) {
vector<pair<int, int>> f = fac.factorize(n);
int a = 0, p = 1;
for (int i = 0; i < ((int)(f.size())); ++i) {
int q = 1;
for (int j = 0; j < f[i].second; ++j) q *= f[i].first;
int b = square_root_prime_modulo(c, q, f[i].first, f[i].second);
if (b == -1) return -1;
a = unite_mod(a, p, b, q);
p *= q;
}
return a;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cerr.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2) && (y1 <= y2))
cout << "Polycarp\n";
else if (x1 + y1 <= max(x2, y2))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
```
|
#include <bits/stdc++.h>
int xg = 0;
int yg = 0;
int dist_p(int x, int y) { return x + y; }
int dist_v(int x, int y) { return std::max(x, y); }
int main(int argc, char** argv) {
int x1 = 0;
int y1 = 0;
int x2 = 0;
int y2 = 0;
std::cin >> x1 >> y1 >> x2 >> y2;
int p_mx[] = {0, -1, 0};
int p_my[] = {0, 0, -1};
int v_mx[] = {0, -1, 0, -1};
int v_my[] = {0, 0, -1, -1};
bool ok = x1 <= x2 && y1 <= y2;
ok = ok || dist_p(x1, y1) <= dist_v(x2, y2);
if (ok) {
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>
int xg = 0;
int yg = 0;
int dist_p(int x, int y) { return x + y; }
int dist_v(int x, int y) { return std::max(x, y); }
int main(int argc, char** argv) {
int x1 = 0;
int y1 = 0;
int x2 = 0;
int y2 = 0;
std::cin >> x1 >> y1 >> x2 >> y2;
int p_mx[] = {0, -1, 0};
int p_my[] = {0, 0, -1};
int v_mx[] = {0, -1, 0, -1};
int v_my[] = {0, 0, -1, -1};
bool ok = x1 <= x2 && y1 <= y2;
ok = ok || dist_p(x1, y1) <= dist_v(x2, y2);
if (ok) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
int tejas_919(int kkkk) {
lli n, m, k, q, u, v, temp = 0, ans = 0;
lli xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
lli mov = 0;
while (xv > 0 || yv > 0) {
mov++;
xv = max(0LL, xv - 1);
yv = max(0LL, yv - 1);
temp = abs(xv - xp) + abs(yv - yp);
{}
if (mov >= temp) return cout << "Polycarp" << '\n', 0;
}
cout << "Vasiliy" << '\n';
return 0;
}
signed main() {
if (!0) {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
}
cout << fixed << setprecision(10);
int t = 1;
for (int i = 0; i < t; i++) {
tejas_919(i + 1);
}
{};
if (0) system("pause");
}
|
### 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;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
int tejas_919(int kkkk) {
lli n, m, k, q, u, v, temp = 0, ans = 0;
lli xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
lli mov = 0;
while (xv > 0 || yv > 0) {
mov++;
xv = max(0LL, xv - 1);
yv = max(0LL, yv - 1);
temp = abs(xv - xp) + abs(yv - yp);
{}
if (mov >= temp) return cout << "Polycarp" << '\n', 0;
}
cout << "Vasiliy" << '\n';
return 0;
}
signed main() {
if (!0) {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
}
cout << fixed << setprecision(10);
int t = 1;
for (int i = 0; i < t; i++) {
tejas_919(i + 1);
}
{};
if (0) system("pause");
}
```
|
#include <bits/stdc++.h>
using namespace std;
int const MAX_N = 5010;
int const MAX_Q = 410100;
int const MAX_CH = 410100;
long long const LL_INF = 1000000000000000009LL;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int go_1 = x1 + y1;
int go_2 = max(x2, y2);
if (go_1 <= 1) {
cout << "Polycarp";
return 0;
}
if (y1 == y2 && x1 == x2 + 1) {
cout << "Vasiliy";
return 0;
}
if (y1 == y2 && x1 == x2 - 1) {
cout << "Polycarp";
return 0;
}
if (x1 == x2 && y1 == y2 + 1) {
cout << "Vasiliy";
return 0;
}
if (x1 == x2 && y1 == y2 - 1) {
cout << "Polycarp";
return 0;
}
if (x2 == x1 - 1 && y2 == y1 + 1) {
if (x2 >= 1)
cout << "Vasiliy";
else
cout << "Polycarp";
return 0;
}
if (x2 == x1 - 1 && y2 == y1 - 1) {
cout << "Vasiliy";
return 0;
}
if (x2 == x1 + 1 && y2 == y1 + 1) {
cout << "Polycarp";
return 0;
}
if (x2 == x1 + 1 && y2 == y1 - 1) {
if (y2 >= 1)
cout << "Vasiliy";
else
cout << "Polycarp";
return 0;
}
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp";
return 0;
}
if (go_1 <= go_2) {
cout << "Polycarp";
return 0;
}
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 const MAX_N = 5010;
int const MAX_Q = 410100;
int const MAX_CH = 410100;
long long const LL_INF = 1000000000000000009LL;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int go_1 = x1 + y1;
int go_2 = max(x2, y2);
if (go_1 <= 1) {
cout << "Polycarp";
return 0;
}
if (y1 == y2 && x1 == x2 + 1) {
cout << "Vasiliy";
return 0;
}
if (y1 == y2 && x1 == x2 - 1) {
cout << "Polycarp";
return 0;
}
if (x1 == x2 && y1 == y2 + 1) {
cout << "Vasiliy";
return 0;
}
if (x1 == x2 && y1 == y2 - 1) {
cout << "Polycarp";
return 0;
}
if (x2 == x1 - 1 && y2 == y1 + 1) {
if (x2 >= 1)
cout << "Vasiliy";
else
cout << "Polycarp";
return 0;
}
if (x2 == x1 - 1 && y2 == y1 - 1) {
cout << "Vasiliy";
return 0;
}
if (x2 == x1 + 1 && y2 == y1 + 1) {
cout << "Polycarp";
return 0;
}
if (x2 == x1 + 1 && y2 == y1 - 1) {
if (y2 >= 1)
cout << "Vasiliy";
else
cout << "Polycarp";
return 0;
}
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp";
return 0;
}
if (go_1 <= go_2) {
cout << "Polycarp";
return 0;
}
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 30, Mod = 1e9 + 7;
const long long SQ = 330;
int main() {
ios::sync_with_stdio(0), cin.tie(0);
long long x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2 && y1 <= y2) || x1 + y1 <= max(x2, y2))
cout << "Polycarp";
else
cout << "Vasiliy";
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;
const long long N = 2e5 + 30, Mod = 1e9 + 7;
const long long SQ = 330;
int main() {
ios::sync_with_stdio(0), cin.tie(0);
long long x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2 && y1 <= y2) || x1 + y1 <= max(x2, y2))
cout << "Polycarp";
else
cout << "Vasiliy";
return (0);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv;
while (cin >> xp >> yp >> xv >> yv) {
int p = xp + yp;
if (p <= max(xv, yv) || (xp <= xv && yp <= yv))
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;
int main() {
int xp, yp, xv, yv;
while (cin >> xp >> yp >> xv >> yv) {
int p = xp + yp;
if (p <= max(xv, yv) || (xp <= xv && yp <= yv))
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void setmax(T& x, T y) {
x = max(x, y);
}
template <typename T>
void setmin(T& x, T y) {
x = min(x, y);
}
template <typename T>
void print(const T& a, ostream& out) {
for (auto i : a) out << i << ' ';
out << endl;
}
const double PI = acos(-1.0);
const long long INF = 1e9 + 47;
const long long LINF = INF * INF;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
int ans1 = 0;
int ans2 = 0;
ans1 = xp + yp;
ans2 = max(xv, yv);
bool win = ans1 <= ans2;
if (xp <= xv && yp <= yv) win = 1;
if (win)
cout << "Polycarp";
else
cout << "Vasiliy";
cout << endl;
cerr << "Time elapsed: " << clock() / (double)CLOCKS_PER_SEC << endl;
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>
void setmax(T& x, T y) {
x = max(x, y);
}
template <typename T>
void setmin(T& x, T y) {
x = min(x, y);
}
template <typename T>
void print(const T& a, ostream& out) {
for (auto i : a) out << i << ' ';
out << endl;
}
const double PI = acos(-1.0);
const long long INF = 1e9 + 47;
const long long LINF = INF * INF;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
int ans1 = 0;
int ans2 = 0;
ans1 = xp + yp;
ans2 = max(xv, yv);
bool win = ans1 <= ans2;
if (xp <= xv && yp <= yv) win = 1;
if (win)
cout << "Polycarp";
else
cout << "Vasiliy";
cout << endl;
cerr << "Time elapsed: " << clock() / (double)CLOCKS_PER_SEC << endl;
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 <= xv && yp <= yv)
puts("Polycarp");
else {
if (xp + yp <= max(xv, yv))
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 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>
int max(int a, int b) { return a > b ? a : b; }
int main() {
int x1, y1, x2, y2;
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
int d;
if (x2 > y2) {
d = x2;
} else {
d = y2;
}
if (d >= x1 + y1) {
printf("Polycarp\n");
} else if (x2 == 0 || y2 == 0) {
if (d >= x1 + y1) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else {
if (x1 - x2 + y1 - y2 > 0) {
printf("Vasiliy\n");
} else {
if (x1 <= x2 && y1 <= y2) {
int h = max(abs(y1 - y2), abs(x1 - x2));
if (x1 - x2 + y1 - y2 <= -h) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else if (x1 > x2 && y1 < y2) {
int h = y2 - y1;
if (x1 - x2 + y1 - y2 <= -h) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else if (x1 < x2 && y1 > y2) {
int h = x2 - x1;
if (x1 - x2 + y1 - y2 <= -h) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else {
printf("Vasiliy\n");
}
}
}
}
|
### 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>
int max(int a, int b) { return a > b ? a : b; }
int main() {
int x1, y1, x2, y2;
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
int d;
if (x2 > y2) {
d = x2;
} else {
d = y2;
}
if (d >= x1 + y1) {
printf("Polycarp\n");
} else if (x2 == 0 || y2 == 0) {
if (d >= x1 + y1) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else {
if (x1 - x2 + y1 - y2 > 0) {
printf("Vasiliy\n");
} else {
if (x1 <= x2 && y1 <= y2) {
int h = max(abs(y1 - y2), abs(x1 - x2));
if (x1 - x2 + y1 - y2 <= -h) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else if (x1 > x2 && y1 < y2) {
int h = y2 - y1;
if (x1 - x2 + y1 - y2 <= -h) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else if (x1 < x2 && y1 > y2) {
int h = x2 - x1;
if (x1 - x2 + y1 - y2 <= -h) {
printf("Polycarp\n");
} else {
printf("Vasiliy\n");
}
} else {
printf("Vasiliy\n");
}
}
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
int a, b, c, d;
cin >> a >> b >> c >> d;
if (a + b <= max(c, d))
cout << "Polycarp";
else if (a <= c && b <= d)
cout << "Polycarp";
else
cout << "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() {
ios::sync_with_stdio(0);
int a, b, c, d;
cin >> a >> b >> c >> d;
if (a + b <= max(c, d))
cout << "Polycarp";
else if (a <= c && b <= d)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int main() {
int tag = 0;
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xp + yp <= max(xv, yv))
tag = 1;
else if (xp <= xv && yp <= yv)
tag = 1;
if (tag)
printf("Polycarp\n");
else
printf("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 xp, yp, xv, yv;
int main() {
int tag = 0;
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xp + yp <= max(xv, yv))
tag = 1;
else if (xp <= xv && yp <= yv)
tag = 1;
if (tag)
printf("Polycarp\n");
else
printf("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long a, b, c, d;
cin >> a >> b >> c >> d;
long long pr, vs;
pr = (a + b);
vs = min(c, d);
int h = c - vs;
int g = d - vs;
vs += h + g;
int count1 = 0;
if (a < c && b <= d) {
count1 = INT_MAX;
} else if (b < d && a <= c) {
count1 = INT_MAX;
}
vs += count1;
if (vs < pr) {
cout << "Vasiliy" << endl;
} else {
cout << "Polycarp" << endl;
}
}
|
### 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() {
long long a, b, c, d;
cin >> a >> b >> c >> d;
long long pr, vs;
pr = (a + b);
vs = min(c, d);
int h = c - vs;
int g = d - vs;
vs += h + g;
int count1 = 0;
if (a < c && b <= d) {
count1 = INT_MAX;
} else if (b < d && a <= c) {
count1 = INT_MAX;
}
vs += count1;
if (vs < pr) {
cout << "Vasiliy" << endl;
} else {
cout << "Polycarp" << endl;
}
}
```
|
#include <bits/stdc++.h>
long long bigmod(long long sonkha, long long ghat, long long vag_const) {
long long vag_shesh = 1;
while (ghat > 0) {
if (ghat % 2 == 1) {
vag_shesh = (vag_shesh * sonkha) % vag_const;
}
ghat /= 2;
sonkha = (sonkha * sonkha) % vag_const;
}
return vag_shesh;
}
long long inverse_mod(long long bivajok, long long vag_const) {
return bigmod(bivajok, vag_const - 2, vag_const);
}
using namespace std;
int main() {
int xp, yp, xv, yv, p, v;
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> xp >> yp >> xv >> yv;
;
p = xp + yp;
v = max(xv, yv);
if (p <= v || (xp <= xv && yp <= yv))
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>
long long bigmod(long long sonkha, long long ghat, long long vag_const) {
long long vag_shesh = 1;
while (ghat > 0) {
if (ghat % 2 == 1) {
vag_shesh = (vag_shesh * sonkha) % vag_const;
}
ghat /= 2;
sonkha = (sonkha * sonkha) % vag_const;
}
return vag_shesh;
}
long long inverse_mod(long long bivajok, long long vag_const) {
return bigmod(bivajok, vag_const - 2, vag_const);
}
using namespace std;
int main() {
int xp, yp, xv, yv, p, v;
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> xp >> yp >> xv >> yv;
;
p = xp + yp;
v = max(xv, yv);
if (p <= v || (xp <= xv && yp <= yv))
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, x, y;
int main() {
scanf("%d%d%d%d", &a, &b, &c, &d);
for (int i = 1;; i++) {
x = max(0, c - 1);
y = max(0, d - 1);
if (a >= x && b >= y) {
if (a - x + b - y <= i) {
printf("Polycarp\n");
return 0;
}
}
c = x;
d = y;
if (c == 0 && d == 0) {
printf("Vasiliy\n");
return 0;
}
}
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 a, b, c, d, x, y;
int main() {
scanf("%d%d%d%d", &a, &b, &c, &d);
for (int i = 1;; i++) {
x = max(0, c - 1);
y = max(0, d - 1);
if (a >= x && b >= y) {
if (a - x + b - y <= i) {
printf("Polycarp\n");
return 0;
}
}
c = x;
d = y;
if (c == 0 && d == 0) {
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;
int d1 = min(x1, x2);
int d2 = min(y1, y2);
bool ft = 0;
if (x1 + y1 <= max(x2, y2)) {
printf("Polycarp\n");
return 0;
}
for (int i = 0; i <= d2; i++) {
int t1 = y1 - i + x1 - d1;
int sx = x2 - d2;
int sy = y2 - i;
int t2 = max(sx, sy);
if (t1 > t2) {
ft = 1;
}
}
for (int i = 0; i <= d1; i++) {
int t1 = x1 - i + y1 - d2;
int sx = y2 - d1;
int sy = x2 - i;
int t2 = max(sx, sy);
if (t1 > t2) {
ft = 1;
}
}
if (!ft)
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
|
### 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 x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int d1 = min(x1, x2);
int d2 = min(y1, y2);
bool ft = 0;
if (x1 + y1 <= max(x2, y2)) {
printf("Polycarp\n");
return 0;
}
for (int i = 0; i <= d2; i++) {
int t1 = y1 - i + x1 - d1;
int sx = x2 - d2;
int sy = y2 - i;
int t2 = max(sx, sy);
if (t1 > t2) {
ft = 1;
}
}
for (int i = 0; i <= d1; i++) {
int t1 = x1 - i + y1 - d2;
int sx = y2 - d1;
int sy = x2 - i;
int t2 = max(sx, sy);
if (t1 > t2) {
ft = 1;
}
}
if (!ft)
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
```
|
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double pii = 2 * pi;
const double eps = 1e-6;
const long long MOD = 1e9 + 7;
void solve() {
int a, b, x, y;
cin >> a >> b >> x >> y;
vector<pair<int, int>> mvs;
while (x != 0 || y != 0) {
mvs.push_back({x, y});
if (x && y) {
x--;
y--;
} else {
if (x)
x--;
else
y--;
}
}
mvs.push_back({0, 0});
bool flag = a + b < mvs.size();
for (int i = 0; i < mvs.size(); i++) {
if (mvs[i].first <= a && mvs[i].second <= b &&
abs(mvs[i].first - a) + abs(mvs[i].second - b) <= i)
flag = true;
}
if (flag)
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
solve();
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;
const double pi = acos(-1.0);
const double pii = 2 * pi;
const double eps = 1e-6;
const long long MOD = 1e9 + 7;
void solve() {
int a, b, x, y;
cin >> a >> b >> x >> y;
vector<pair<int, int>> mvs;
while (x != 0 || y != 0) {
mvs.push_back({x, y});
if (x && y) {
x--;
y--;
} else {
if (x)
x--;
else
y--;
}
}
mvs.push_back({0, 0});
bool flag = a + b < mvs.size();
for (int i = 0; i < mvs.size(); i++) {
if (mvs[i].first <= a && mvs[i].second <= b &&
abs(mvs[i].first - a) + abs(mvs[i].second - b) <= i)
flag = true;
}
if (flag)
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
solve();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int p_x, p_y, v_x, v_y;
void vas_move() {
if (v_x > 0 and v_y > 0 and
make_pair(p_x, p_y) != make_pair(v_x - 1, v_y - 1)) {
v_x--;
v_y--;
} else if (v_x > v_y) {
if (make_pair(v_x - 1, v_y) == make_pair(p_x, p_y)) {
if (v_y != 0) v_y--;
} else if (v_x > 0)
v_x--;
} else if (v_x <= v_y) {
if (make_pair(v_x, v_y - 1) == make_pair(p_x, p_y)) {
if (v_x != 0) v_x--;
} else if (v_y > 0)
v_y--;
}
if (v_x == 0 and v_y == 0) {
cout << "Vasiliy";
exit(0);
}
}
long long manh_dist(long long A, long long B, long long C, long long D) {
return abs(A - C) + abs(B - D);
}
void pol_move() {
if ((p_x - 1 == v_x and p_y == v_y) or p_x == 0)
p_y--;
else if ((p_x == v_x and p_y - 1 == v_y) or p_y == 0)
p_x--;
else if (p_x - v_x > p_y - v_y)
p_x--;
else
p_y--;
if (p_x == 0 and p_y == 0) {
cout << "Polycarp";
exit(0);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> p_x >> p_y >> v_x >> v_y;
int it = 0;
while (1) {
pol_move();
vas_move();
}
}
|
### 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 p_x, p_y, v_x, v_y;
void vas_move() {
if (v_x > 0 and v_y > 0 and
make_pair(p_x, p_y) != make_pair(v_x - 1, v_y - 1)) {
v_x--;
v_y--;
} else if (v_x > v_y) {
if (make_pair(v_x - 1, v_y) == make_pair(p_x, p_y)) {
if (v_y != 0) v_y--;
} else if (v_x > 0)
v_x--;
} else if (v_x <= v_y) {
if (make_pair(v_x, v_y - 1) == make_pair(p_x, p_y)) {
if (v_x != 0) v_x--;
} else if (v_y > 0)
v_y--;
}
if (v_x == 0 and v_y == 0) {
cout << "Vasiliy";
exit(0);
}
}
long long manh_dist(long long A, long long B, long long C, long long D) {
return abs(A - C) + abs(B - D);
}
void pol_move() {
if ((p_x - 1 == v_x and p_y == v_y) or p_x == 0)
p_y--;
else if ((p_x == v_x and p_y - 1 == v_y) or p_y == 0)
p_x--;
else if (p_x - v_x > p_y - v_y)
p_x--;
else
p_y--;
if (p_x == 0 and p_y == 0) {
cout << "Polycarp";
exit(0);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> p_x >> p_y >> v_x >> v_y;
int it = 0;
while (1) {
pol_move();
vas_move();
}
}
```
|
#include <bits/stdc++.h>
int xp, yp, xv, yv;
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xp + yp <= yv || xp + yp <= xv || (xp <= xv && yp <= yv))
printf("Polycarp");
else
printf("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>
int xp, yp, xv, yv;
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xp + yp <= yv || xp + yp <= xv || (xp <= xv && yp <= yv))
printf("Polycarp");
else
printf("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long a, b, x, y;
int main() {
cin >> a >> b >> x >> y;
if (x + y < a + b) {
cout << "Vasiliy" << endl;
} else if (x >= a and y >= b) {
cout << "Polycarp" << endl;
} else if (max(x, y) >= a + b) {
cout << "Polycarp" << endl;
} else
cout << "Vasiliy" << endl;
}
|
### 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;
long long a, b, x, y;
int main() {
cin >> a >> b >> x >> y;
if (x + y < a + b) {
cout << "Vasiliy" << endl;
} else if (x >= a and y >= b) {
cout << "Polycarp" << endl;
} else if (max(x, y) >= a + b) {
cout << "Polycarp" << endl;
} else
cout << "Vasiliy" << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xv + yv < xp + yp) {
cout << "Vasiliy\n";
return 0;
}
if (xp <= xv && yp <= yv) {
cout << "Polycarp\n";
return 0;
}
if (max(xv, yv) >= (xp + yp))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
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 xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xv + yv < xp + yp) {
cout << "Vasiliy\n";
return 0;
}
if (xp <= xv && yp <= yv) {
cout << "Polycarp\n";
return 0;
}
if (max(xv, yv) >= (xp + yp))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
```
|
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp + yp <= max(xv, yv) || (xp <= xv && yp <= yv))
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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp + yp <= max(xv, yv) || (xp <= xv && yp <= yv))
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int x, y, i, j, dist;
int main() {
cin >> x >> y >> i >> j;
cout << (x + y <= max(i, j) || (x <= i && y <= j) ? "Polycarp\n"
: "Vasiliy\n");
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 x, y, i, j, dist;
int main() {
cin >> x >> y >> i >> j;
cout << (x + y <= max(i, j) || (x <= i && y <= j) ? "Polycarp\n"
: "Vasiliy\n");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp" << endl;
} else 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, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp" << endl;
} else if (x1 + y1 <= max(x2, y2)) {
cout << "Polycarp" << endl;
} else {
cout << "Vasiliy" << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp <= xv && yp <= yv) {
cout << "Polycarp\n";
} else if (xp >= xv && yp >= yv) {
cout << "Vasiliy\n";
} else {
if (yp < yv) {
swap(xp, yp), swap(xv, yv);
}
cout << (yp - (xv - xp) <= max(0, yv - (xv - xp)) ? "Polycarp" : "Vasiliy")
<< "\n";
}
}
|
### 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() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp <= xv && yp <= yv) {
cout << "Polycarp\n";
} else if (xp >= xv && yp >= yv) {
cout << "Vasiliy\n";
} else {
if (yp < yv) {
swap(xp, yp), swap(xv, yv);
}
cout << (yp - (xv - xp) <= max(0, yv - (xv - xp)) ? "Polycarp" : "Vasiliy")
<< "\n";
}
}
```
|
#include <bits/stdc++.h>
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
puts(d >= a + b || c >= a + b || (c >= a && d >= b) ? "Polycarp" : "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>
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
puts(d >= a + b || c >= a + b || (c >= a && d >= b) ? "Polycarp" : "Vasiliy");
}
```
|
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long x[2], y[2], s1, s2;
int32_t main() {
cin >> x[0] >> y[0];
cin >> x[1] >> y[1];
s1 = min(x[1], y[1]) + max(x[1], y[1]) - min(x[1], y[1]);
s2 = x[0] + y[0];
if (s2 <= s1 || (x[0] <= x[1] && y[0] <= y[1])) {
cout << "Polycarp";
} else
cout << "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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long x[2], y[2], s1, s2;
int32_t main() {
cin >> x[0] >> y[0];
cin >> x[1] >> y[1];
s1 = min(x[1], y[1]) + max(x[1], y[1]) - min(x[1], y[1]);
s2 = x[0] + y[0];
if (s2 <= s1 || (x[0] <= x[1] && y[0] <= y[1])) {
cout << "Polycarp";
} else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
struct node {
int n;
int at;
};
bool operator<(node a, node b) { return a.at > b.at; }
void preprocess() {}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.precision(17);
int px, py, vx, vy;
cin >> px >> py >> vx >> vy;
int p = max(px, py), v = max(vx, vy);
if (px <= vx && py <= vy)
cout << "Polycarp";
else if (v >= px + py)
cout << "Polycarp";
else
cout << "Vasiliy";
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;
struct node {
int n;
int at;
};
bool operator<(node a, node b) { return a.at > b.at; }
void preprocess() {}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.precision(17);
int px, py, vx, vy;
cin >> px >> py >> vx >> vy;
int p = max(px, py), v = max(vx, vy);
if (px <= vx && py <= vy)
cout << "Polycarp";
else if (v >= px + py)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
const int MAX_LEN = 1010;
using namespace std;
template <typename U, typename V>
string to_string(pair<U, V>);
string to_string(const string& e_) { return "\"" + e_ + "\""; }
string to_string(char e_) { return "\'" + string(1, e_) + "\'"; }
string to_string(bool e_) { return e_ ? "true" : "false"; }
template <typename T>
string to_string(T e_) {
string s_ = "[ ";
for (const auto& x_ : e_) s_ += to_string(x_) + " ";
return s_ + "]";
}
template <typename U, typename V>
string to_string(pair<U, V> e_) {
return "(" + to_string(e_.first) + ", " + to_string(e_.second) + ")";
}
void dbg_str() { ; }
template <typename U, typename... V>
void dbg_str(U u, V... v) {
;
dbg_str(v...);
}
long long int xp, yp, xv, yv;
int main() {
scanf("%lld %lld %lld %lld", &xp, &yp, &xv, &yv);
if (xp + yp <= (max(xv, yv)) || (xp <= xv && yp <= yv))
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
|
### 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>
const int MAX_LEN = 1010;
using namespace std;
template <typename U, typename V>
string to_string(pair<U, V>);
string to_string(const string& e_) { return "\"" + e_ + "\""; }
string to_string(char e_) { return "\'" + string(1, e_) + "\'"; }
string to_string(bool e_) { return e_ ? "true" : "false"; }
template <typename T>
string to_string(T e_) {
string s_ = "[ ";
for (const auto& x_ : e_) s_ += to_string(x_) + " ";
return s_ + "]";
}
template <typename U, typename V>
string to_string(pair<U, V> e_) {
return "(" + to_string(e_.first) + ", " + to_string(e_.second) + ")";
}
void dbg_str() { ; }
template <typename U, typename... V>
void dbg_str(U u, V... v) {
;
dbg_str(v...);
}
long long int xp, yp, xv, yv;
int main() {
scanf("%lld %lld %lld %lld", &xp, &yp, &xv, &yv);
if (xp + yp <= (max(xv, yv)) || (xp <= xv && yp <= yv))
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
string a = "Polycarp", b = "Vasiliy";
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 and y1 <= y2) {
cout << a << "\n";
return 0;
}
if (x1 >= x2 and y1 >= y2) {
cout << b;
return 0;
}
if (x1 + y1 <= max(x2, y2))
cout << a;
else
cout << b;
}
|
### 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() {
string a = "Polycarp", b = "Vasiliy";
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 and y1 <= y2) {
cout << a << "\n";
return 0;
}
if (x1 >= x2 and y1 >= y2) {
cout << b;
return 0;
}
if (x1 + y1 <= max(x2, y2))
cout << a;
else
cout << b;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 2;
const double EPS = 1e-11;
const long long oo = 1e16;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int fin = 0;
while (fin == 0) {
if (x1 + y1 <= max(x2, y2)) {
fin = 1;
break;
}
if (x1 - y1 > x2 - y2) {
if (x1 >= 1)
x1--;
else
y1--;
} else if (x1 - y1 < x2 - y2) {
if (y1 >= 1)
y1--;
else
x1--;
} else {
if (x1 >= 1)
x1--;
else
y1--;
}
if (x1 == 0 && y1 == 0) {
fin = 1;
break;
}
if ((x2 - 1 != x1 || y2 - 1 != y1) && x2 >= 1 && y2 >= 1) {
x2--;
y2--;
} else if ((x2 != x1 || y2 - 1 != y1) && y2 >= 1) {
y2--;
} else if ((x2 - 1 != x1 || y2 != y1) && x2 >= 1) {
x2--;
}
if (x2 == 0 && y2 == 0) {
fin = 2;
break;
}
}
if (fin == 1)
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
|
### 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;
const int MAX = 2e5 + 2;
const double EPS = 1e-11;
const long long oo = 1e16;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int fin = 0;
while (fin == 0) {
if (x1 + y1 <= max(x2, y2)) {
fin = 1;
break;
}
if (x1 - y1 > x2 - y2) {
if (x1 >= 1)
x1--;
else
y1--;
} else if (x1 - y1 < x2 - y2) {
if (y1 >= 1)
y1--;
else
x1--;
} else {
if (x1 >= 1)
x1--;
else
y1--;
}
if (x1 == 0 && y1 == 0) {
fin = 1;
break;
}
if ((x2 - 1 != x1 || y2 - 1 != y1) && x2 >= 1 && y2 >= 1) {
x2--;
y2--;
} else if ((x2 != x1 || y2 - 1 != y1) && y2 >= 1) {
y2--;
} else if ((x2 - 1 != x1 || y2 != y1) && x2 >= 1) {
x2--;
}
if (x2 == 0 && y2 == 0) {
fin = 2;
break;
}
}
if (fin == 1)
printf("Polycarp\n");
else
printf("Vasiliy\n");
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) {
if (a > b) {
return a;
} else {
return b;
}
}
long long min(long long a, long long b) {
if (a < b) {
return a;
} else {
return b;
}
}
struct P {
double x, y;
P() {}
P(int x, int y) : x(double(x)), y(double(y)) {}
P(double x, double y) : x(x), y(y) {}
P operator+(const P &a) const { return P(x + a.x, y + a.y); }
P operator-(const P &a) const { return P(x - a.x, y - a.y); }
double operator^(const P &a) const { return x * a.x + y * a.y; }
void in() { scanf("%lf%lf", &x, &y); }
void out() { printf("REQUIRED %.7f %.7f\n", x, y); }
double dist(P a, P b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
double sqdist(P a, P b) {
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
};
int xp, yp, xv, yv;
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xv >= xp && yv >= yp || (xp + yp <= max(xv, yv)))
puts("Polycarp");
else
puts("Vasiliy");
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;
long long max(long long a, long long b) {
if (a > b) {
return a;
} else {
return b;
}
}
long long min(long long a, long long b) {
if (a < b) {
return a;
} else {
return b;
}
}
struct P {
double x, y;
P() {}
P(int x, int y) : x(double(x)), y(double(y)) {}
P(double x, double y) : x(x), y(y) {}
P operator+(const P &a) const { return P(x + a.x, y + a.y); }
P operator-(const P &a) const { return P(x - a.x, y - a.y); }
double operator^(const P &a) const { return x * a.x + y * a.y; }
void in() { scanf("%lf%lf", &x, &y); }
void out() { printf("REQUIRED %.7f %.7f\n", x, y); }
double dist(P a, P b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
double sqdist(P a, P b) {
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
};
int xp, yp, xv, yv;
int main() {
scanf("%d%d%d%d", &xp, &yp, &xv, &yv);
if (xv >= xp && yv >= yp || (xp + yp <= max(xv, yv)))
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2))
cout << "Polycarp";
else if (x1 <= x2 && y1 <= y2)
cout << "Polycarp";
else
cout << "Vasiliy";
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() {
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2))
cout << "Polycarp";
else if (x1 <= x2 && y1 <= y2)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
bool solve(long long x1, long long y1, long long x2, long long y2) {
if (x1 <= x2 && y1 <= y2) return true;
if (y2 < y1 && x2 < x1) return false;
if (y2 < y1) {
swap(x1, y1);
swap(x2, y2);
}
long long d1 = x1;
long long d2 = max(x2, y2 - y1);
return d1 <= d2;
}
int main() {
long long x1, y1;
scanf("%lld%lld", &x1, &y1);
long long x2, y2;
scanf("%lld%lld", &x2, &y2);
puts(solve(x1, y1, x2, y2) ? "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;
bool solve(long long x1, long long y1, long long x2, long long y2) {
if (x1 <= x2 && y1 <= y2) return true;
if (y2 < y1 && x2 < x1) return false;
if (y2 < y1) {
swap(x1, y1);
swap(x2, y2);
}
long long d1 = x1;
long long d2 = max(x2, y2 - y1);
return d1 <= d2;
}
int main() {
long long x1, y1;
scanf("%lld%lld", &x1, &y1);
long long x2, y2;
scanf("%lld%lld", &x2, &y2);
puts(solve(x1, y1, x2, y2) ? "Polycarp" : "Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2000 * 100 + 12;
int main() {
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
int ki = 1;
while (x1 || x2 || y1 || y2) {
if (ki % 2) {
if (x1 && y1) {
if (y2 > y1)
x1--;
else
y1--;
} else if (x1)
x1--;
else
y1--;
} else {
if (x2 && y2) {
if (x2 != x1 + 1 || y2 != y1 + 1)
x2--, y2--;
else
x2--;
} else if (x2)
x2--;
else
y2--;
}
if (x1 == 0 && y1 == 0) {
cout << "Polycarp";
return 0;
}
if (x2 == 0 && y2 == 0) {
cout << "Vasiliy";
return 0;
}
ki++;
}
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 maxn = 2000 * 100 + 12;
int main() {
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
int ki = 1;
while (x1 || x2 || y1 || y2) {
if (ki % 2) {
if (x1 && y1) {
if (y2 > y1)
x1--;
else
y1--;
} else if (x1)
x1--;
else
y1--;
} else {
if (x2 && y2) {
if (x2 != x1 + 1 || y2 != y1 + 1)
x2--, y2--;
else
x2--;
} else if (x2)
x2--;
else
y2--;
}
if (x1 == 0 && y1 == 0) {
cout << "Polycarp";
return 0;
}
if (x2 == 0 && y2 == 0) {
cout << "Vasiliy";
return 0;
}
ki++;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
inline int manh(int x1, int y1, int x2, int y2) {
return abs(x1 - x2) + abs(y1 - y2);
}
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int k = min(x2, y2);
int xx2 = x2 - k;
int yy2 = y2 - k;
bool can = 0;
if (manh(0, 0, x1, y1) <= max(x2, y2)) can = 1;
for (int i = xx2, j = yy2; !can && i <= x2; ++i, ++j)
if (manh(i, j, x1, y1) <= x2 - i) can = 1;
if (can)
printf("Polycarp");
else
printf("Vasiliy");
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;
inline int manh(int x1, int y1, int x2, int y2) {
return abs(x1 - x2) + abs(y1 - y2);
}
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int k = min(x2, y2);
int xx2 = x2 - k;
int yy2 = y2 - k;
bool can = 0;
if (manh(0, 0, x1, y1) <= max(x2, y2)) can = 1;
for (int i = xx2, j = yy2; !can && i <= x2; ++i, ++j)
if (manh(i, j, x1, y1) <= x2 - i) can = 1;
if (can)
printf("Polycarp");
else
printf("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
void Vasiliy() {
cout << "Vasiliy" << endl;
exit(0);
}
void Polycarp() {
cout << "Polycarp" << endl;
exit(0);
}
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp <= xv && yp <= yv) Polycarp();
if (xv < xp && yv < yp) Vasiliy();
if (xv < xp) {
if (yv - yp < xp)
Vasiliy();
else
Polycarp();
}
if (yv < yp) {
if (xv - xp < yp)
Vasiliy();
else
Polycarp();
}
}
|
### 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 Vasiliy() {
cout << "Vasiliy" << endl;
exit(0);
}
void Polycarp() {
cout << "Polycarp" << endl;
exit(0);
}
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp <= xv && yp <= yv) Polycarp();
if (xv < xp && yv < yp) Vasiliy();
if (xv < xp) {
if (yv - yp < xp)
Vasiliy();
else
Polycarp();
}
if (yv < yp) {
if (xv - xp < yp)
Vasiliy();
else
Polycarp();
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int x, y, x1, y11;
void get() {
int t = x1 - x;
if (t <= 0) {
return;
}
int v = max(0, y11 - t);
int t1 = y - v;
if (t1 <= t) {
printf("Polycarp\n");
exit(0);
}
}
int main() {
cin >> x >> y >> x1 >> y11;
get();
swap(x, y);
swap(x1, y11);
get();
printf("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 x, y, x1, y11;
void get() {
int t = x1 - x;
if (t <= 0) {
return;
}
int v = max(0, y11 - t);
int t1 = y - v;
if (t1 <= t) {
printf("Polycarp\n");
exit(0);
}
}
int main() {
cin >> x >> y >> x1 >> y11;
get();
swap(x, y);
swap(x1, y11);
get();
printf("Vasiliy\n");
return 0;
}
```
|
#include <bits/stdc++.h>
const int INF = 1e9;
const int N = 312312;
const double PI = 3.1415926535897932384626433832795;
using namespace std;
int main() {
ios_base ::sync_with_stdio();
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2))
cout << "Polycarp";
else if (x1 <= x2 && y1 <= y2)
cout << "Polycarp";
else
cout << "Vasiliy" << endl;
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>
const int INF = 1e9;
const int N = 312312;
const double PI = 3.1415926535897932384626433832795;
using namespace std;
int main() {
ios_base ::sync_with_stdio();
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 + y1 <= max(x2, y2))
cout << "Polycarp";
else if (x1 <= x2 && y1 <= y2)
cout << "Polycarp";
else
cout << "Vasiliy" << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
int win = 0;
while (cin >> x1 >> y1 >> x2 >> y2) {
if (x1 <= x2 && y1 <= y2) win = 1;
if (x1 + y1 <= max(x2, y2)) win = 1;
puts(win == 1 ? "Polycarp" : "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;
int win = 0;
while (cin >> x1 >> y1 >> x2 >> y2) {
if (x1 <= x2 && y1 <= y2) win = 1;
if (x1 + y1 <= max(x2, y2)) win = 1;
puts(win == 1 ? "Polycarp" : "Vasiliy");
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int ax, ay, bx, by;
scanf("%d%d%d%d", &ax, &ay, &bx, &by);
if (ax <= bx && ay <= by) {
printf("Polycarp\n");
return 0;
}
if (ax + ay <= bx + by - min(bx, by)) {
printf("Polycarp\n");
} else
printf("Vasiliy\n");
}
|
### 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 ax, ay, bx, by;
scanf("%d%d%d%d", &ax, &ay, &bx, &by);
if (ax <= bx && ay <= by) {
printf("Polycarp\n");
return 0;
}
if (ax + ay <= bx + by - min(bx, by)) {
printf("Polycarp\n");
} else
printf("Vasiliy\n");
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp <= xv && yp <= yv)
cout << "Polycarp" << endl;
else if (max(xv, yv) < xp + yp)
cout << "Vasiliy" << endl;
else
cout << "Polycarp" << endl;
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 xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp <= xv && yp <= yv)
cout << "Polycarp" << endl;
else if (max(xv, yv) < xp + yp)
cout << "Vasiliy" << endl;
else
cout << "Polycarp" << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int x, y, x2, y2;
int main() {
cin >> x >> y >> x2 >> y2;
if (x <= x2 && y <= y2 || x + y <= max(x2, y2)) {
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;
int x, y, x2, y2;
int main() {
cin >> x >> y >> x2 >> y2;
if (x <= x2 && y <= y2 || x + y <= max(x2, y2)) {
cout << "Polycarp\n";
} else {
cout << "Vasiliy\n";
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int go1 = x1 + y1;
int mn = min(x2, y2);
int go2 = mn + (x2 + y2 - 2 * mn);
if (go1 <= go2) {
cout << "Polycarp";
exit(0);
}
for (int i = mn; i >= 0; i--) {
int go = mn - i;
int gox = x2 - go, goy = y2 - go;
if (abs(gox - x1) + abs(goy - y1) <= go) {
cout << "Polycarp";
exit(0);
}
}
cout << "Vasiliy";
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() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int go1 = x1 + y1;
int mn = min(x2, y2);
int go2 = mn + (x2 + y2 - 2 * mn);
if (go1 <= go2) {
cout << "Polycarp";
exit(0);
}
for (int i = mn; i >= 0; i--) {
int go = mn - i;
int gox = x2 - go, goy = y2 - go;
if (abs(gox - x1) + abs(goy - y1) <= go) {
cout << "Polycarp";
exit(0);
}
}
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
const long long inf = 1000000000ll;
const long long inf64 = inf * inf;
const long long base = inf + 7;
const long long MOD = inf + 9;
const double pi = acos(-1.0);
using namespace std;
bool solve() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp == xv) {
if (yp < yv)
puts("Polycarp");
else
puts("Vasiliy");
} else if (yp == yv) {
if (xp < xv)
puts("Polycarp");
else
puts("Vasiliy");
} else if (xp < xv && yp < yv)
puts("Polycarp");
else if (xv < xp && yv < yp)
puts("Vasiliy");
else {
int sp = xp + yp;
int sv = max(xv, yv);
if (sp <= sv)
puts("Polycarp");
else
puts("Vasiliy");
}
return true;
}
int main() {
solve();
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>
const long long inf = 1000000000ll;
const long long inf64 = inf * inf;
const long long base = inf + 7;
const long long MOD = inf + 9;
const double pi = acos(-1.0);
using namespace std;
bool solve() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp == xv) {
if (yp < yv)
puts("Polycarp");
else
puts("Vasiliy");
} else if (yp == yv) {
if (xp < xv)
puts("Polycarp");
else
puts("Vasiliy");
} else if (xp < xv && yp < yv)
puts("Polycarp");
else if (xv < xp && yv < yp)
puts("Vasiliy");
else {
int sp = xp + yp;
int sv = max(xv, yv);
if (sp <= sv)
puts("Polycarp");
else
puts("Vasiliy");
}
return true;
}
int main() {
solve();
return 0;
}
```
|
#include <bits/stdc++.h>
inline int max(int a, int b) { return a > b ? a : b; }
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
if ((a <= c && b <= d) || (a + b <= max(c, d)))
printf("Polycarp\n");
else
printf("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>
inline int max(int a, int b) { return a > b ? a : b; }
int main() {
int a, b, c, d;
scanf("%d%d%d%d", &a, &b, &c, &d);
if ((a <= c && b <= d) || (a + b <= max(c, d)))
printf("Polycarp\n");
else
printf("Vasiliy\n");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int L = 100;
bool f[L + 1][L + 1][L + 1][L + 1];
bool vis[L + 1][L + 1][L + 1][L + 1];
bool dfs(int x1, int y1, int x2, int y2) {
if (x1 < 0 || y1 < 0) return false;
if (x2 < 0 || y2 < 0) return true;
if (x1 == 0 && y1 == 0) return true;
if (x2 == 0 && y2 == 0) return false;
if (vis[x1][y1][x2][y2]) return f[x1][y1][x2][y2];
vis[x1][y1][x2][y2] = true;
for (int dx = 0; dx <= 1; dx++)
for (int dy = 0; dy <= 1; dy++)
if (dx + dy <= 1) {
if (x1 - dx == x2 && y1 - dy == y2) continue;
bool b = true;
for (int cx = 0; cx <= 1; cx++)
for (int cy = 0; cy <= 1; cy++) {
if (x1 - dx == x2 - cx && y1 - dy == y2 - cy) continue;
if (!dfs(x1 - dx, y1 - dy, x2 - cx, y2 - cy)) b = false;
}
if (b == true) return f[x1][y1][x2][y2] = true;
}
return f[x1][y1][x2][y2] = false;
}
inline bool solve(int x1, int y1, int x2, int y2) {
if (x1 <= x2 && y1 <= y2) return true;
return x1 + y1 <= max(x2, y2);
}
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
puts(solve(x1, y1, x2, y2) ? "Polycarp" : "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;
const int L = 100;
bool f[L + 1][L + 1][L + 1][L + 1];
bool vis[L + 1][L + 1][L + 1][L + 1];
bool dfs(int x1, int y1, int x2, int y2) {
if (x1 < 0 || y1 < 0) return false;
if (x2 < 0 || y2 < 0) return true;
if (x1 == 0 && y1 == 0) return true;
if (x2 == 0 && y2 == 0) return false;
if (vis[x1][y1][x2][y2]) return f[x1][y1][x2][y2];
vis[x1][y1][x2][y2] = true;
for (int dx = 0; dx <= 1; dx++)
for (int dy = 0; dy <= 1; dy++)
if (dx + dy <= 1) {
if (x1 - dx == x2 && y1 - dy == y2) continue;
bool b = true;
for (int cx = 0; cx <= 1; cx++)
for (int cy = 0; cy <= 1; cy++) {
if (x1 - dx == x2 - cx && y1 - dy == y2 - cy) continue;
if (!dfs(x1 - dx, y1 - dy, x2 - cx, y2 - cy)) b = false;
}
if (b == true) return f[x1][y1][x2][y2] = true;
}
return f[x1][y1][x2][y2] = false;
}
inline bool solve(int x1, int y1, int x2, int y2) {
if (x1 <= x2 && y1 <= y2) return true;
return x1 + y1 <= max(x2, y2);
}
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
puts(solve(x1, y1, x2, y2) ? "Polycarp" : "Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int x, y, X, Y;
int win() {
if (X < Y) {
swap(x, y);
swap(X, Y);
}
if (x <= X && y <= Y) return 1;
if (x > X) return 2;
if (y <= X - x) return 1;
return 2;
}
int main() {
scanf("%d %d %d %d", &x, &y, &X, &Y);
puts(win() == 1 ? "Polycarp" : "Vasiliy");
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 x, y, X, Y;
int win() {
if (X < Y) {
swap(x, y);
swap(X, Y);
}
if (x <= X && y <= Y) return 1;
if (x > X) return 2;
if (y <= X - x) return 1;
return 2;
}
int main() {
scanf("%d %d %d %d", &x, &y, &X, &Y);
puts(win() == 1 ? "Polycarp" : "Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int x, y, x2, y2;
int main() {
ios_base::sync_with_stdio(0);
string poly = "Polycarp\n", vas = "Vasiliy\n";
cin >> x >> y >> x2 >> y2;
if (x <= x2 && y <= y2) {
cout << poly;
return 0;
}
int a = x + y, b = max(x2, y2);
if (a <= b) {
cout << poly;
return 0;
}
cout << vas;
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 x, y, x2, y2;
int main() {
ios_base::sync_with_stdio(0);
string poly = "Polycarp\n", vas = "Vasiliy\n";
cin >> x >> y >> x2 >> y2;
if (x <= x2 && y <= y2) {
cout << poly;
return 0;
}
int a = x + y, b = max(x2, y2);
if (a <= b) {
cout << poly;
return 0;
}
cout << vas;
return 0;
}
```
|
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:134217728")
const long long MOD = 1000000007;
const int INF = 1000000000;
const int MAXN = 200005;
const double EPS = 1e-6;
const int HASH_POW = 7;
const double PI = acos(-1.0);
using namespace std;
void my_return(int code) { exit(code); }
int main() {
mt19937 mt_rand(time(NULL));
int x1, y1, x2, y2;
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
if (x1 + y1 <= max(x2, y2))
printf("Polycarp\n");
else if (x1 <= x2 && y1 <= y2)
printf("Polycarp\n");
else
printf("Vasiliy\n");
my_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>
#pragma comment(linker, "/STACK:134217728")
const long long MOD = 1000000007;
const int INF = 1000000000;
const int MAXN = 200005;
const double EPS = 1e-6;
const int HASH_POW = 7;
const double PI = acos(-1.0);
using namespace std;
void my_return(int code) { exit(code); }
int main() {
mt19937 mt_rand(time(NULL));
int x1, y1, x2, y2;
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
if (x1 + y1 <= max(x2, y2))
printf("Polycarp\n");
else if (x1 <= x2 && y1 <= y2)
printf("Polycarp\n");
else
printf("Vasiliy\n");
my_return(0);
}
```
|
#include <bits/stdc++.h>
using namespace std;
template <typename T>
string toStr(T Number) {
stringstream ss;
ss << Number;
return ss.str();
}
template <typename T>
T toNum(const string &Text) {
stringstream ss(Text);
T result;
return ss >> result ? result : 0;
}
int inGrid(int x, int y, int r, int c) {
return (x >= 0 && x < r && y >= 0 && y < c ? 1 : 0);
}
int leap(int m) {
if (m % 4 == 0 && m % 100 != 0)
return 1;
else if (m % 400 == 0)
return 1;
else
return 0;
}
int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
string months[] = {"January", "February", "March", "April",
"May", "June", "July", "August",
"September", "October", "November", "December"};
int dx4[] = {-1, 0, 1, 0};
int dy4[] = {0, 1, 0, -1};
const double pi = acos(-1.0), eps = 1e-9;
const int N = 2e5 + 5, M = 1e9 + 7;
int xp, yp, xv, yv;
int main() {
cin >> xp >> yp >> xv >> yv;
int xx = xv, yy = yv;
vector<pair<int, int> > v;
v.push_back(make_pair(xv, yv));
for (;;) {
if (xv && yv)
v.push_back(make_pair(--xv, --yv));
else if (xv)
v.push_back(make_pair(--xv, yv));
else if (yv)
v.push_back(make_pair(xv, --yv));
else
break;
}
int f1 = 0, f2 = 0;
for (int i = 0; i < v.size(); i++) {
int c = abs(v[i].first - xp) + abs(v[i].second - yp);
if (c <= i) {
f1 = 1;
break;
}
}
v.clear();
xv = xx, yv = yy;
v.push_back(make_pair(xv, yv));
for (;;) {
if (xv && yv)
v.push_back(make_pair(--xv, --yv));
else if (yv)
v.push_back(make_pair(xv, --yv));
else if (xv)
v.push_back(make_pair(--xv, yv));
else
break;
}
for (int i = 0; i < v.size(); i++) {
int c = abs(v[i].first - xp) + abs(v[i].second - yp);
if (c <= i) {
f2 = 1;
break;
}
}
if (f1 && f2)
puts("Polycarp");
else
puts("Vasiliy");
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;
template <typename T>
string toStr(T Number) {
stringstream ss;
ss << Number;
return ss.str();
}
template <typename T>
T toNum(const string &Text) {
stringstream ss(Text);
T result;
return ss >> result ? result : 0;
}
int inGrid(int x, int y, int r, int c) {
return (x >= 0 && x < r && y >= 0 && y < c ? 1 : 0);
}
int leap(int m) {
if (m % 4 == 0 && m % 100 != 0)
return 1;
else if (m % 400 == 0)
return 1;
else
return 0;
}
int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
string months[] = {"January", "February", "March", "April",
"May", "June", "July", "August",
"September", "October", "November", "December"};
int dx4[] = {-1, 0, 1, 0};
int dy4[] = {0, 1, 0, -1};
const double pi = acos(-1.0), eps = 1e-9;
const int N = 2e5 + 5, M = 1e9 + 7;
int xp, yp, xv, yv;
int main() {
cin >> xp >> yp >> xv >> yv;
int xx = xv, yy = yv;
vector<pair<int, int> > v;
v.push_back(make_pair(xv, yv));
for (;;) {
if (xv && yv)
v.push_back(make_pair(--xv, --yv));
else if (xv)
v.push_back(make_pair(--xv, yv));
else if (yv)
v.push_back(make_pair(xv, --yv));
else
break;
}
int f1 = 0, f2 = 0;
for (int i = 0; i < v.size(); i++) {
int c = abs(v[i].first - xp) + abs(v[i].second - yp);
if (c <= i) {
f1 = 1;
break;
}
}
v.clear();
xv = xx, yv = yy;
v.push_back(make_pair(xv, yv));
for (;;) {
if (xv && yv)
v.push_back(make_pair(--xv, --yv));
else if (yv)
v.push_back(make_pair(xv, --yv));
else if (xv)
v.push_back(make_pair(--xv, yv));
else
break;
}
for (int i = 0; i < v.size(); i++) {
int c = abs(v[i].first - xp) + abs(v[i].second - yp);
if (c <= i) {
f2 = 1;
break;
}
}
if (f1 && f2)
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const long long N = (long long)(5e5) + 322;
const long long INF = (long long)(1e9);
const long long mod = (long long)(1e9) + 7;
const double eps = 1e-9;
int xp, yp, xv, yv, distp, distv;
int main() {
ios_base ::sync_with_stdio(false);
cin.tie(0);
cin >> xp >> yp >> xv >> yv;
distp = xp + yp;
distv = max(xv, yv);
if (distp <= distv) {
cout << "Polycarp";
return 0;
}
if (xp <= xv && yp <= yv) {
cout << "Polycarp";
return 0;
}
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;
const long long N = (long long)(5e5) + 322;
const long long INF = (long long)(1e9);
const long long mod = (long long)(1e9) + 7;
const double eps = 1e-9;
int xp, yp, xv, yv, distp, distv;
int main() {
ios_base ::sync_with_stdio(false);
cin.tie(0);
cin >> xp >> yp >> xv >> yv;
distp = xp + yp;
distv = max(xv, yv);
if (distp <= distv) {
cout << "Polycarp";
return 0;
}
if (xp <= xv && yp <= yv) {
cout << "Polycarp";
return 0;
}
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
bool sol;
if (xp <= xv and yp <= yv) {
sol = true;
} else if (xv <= xp and yv <= yp) {
sol = false;
} else {
int dp = xp + yp;
int dv = max(xv, yv);
sol = dp <= dv;
}
cout << (sol ? "Polycarp" : "Vasiliy") << endl;
}
|
### 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 xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
bool sol;
if (xp <= xv and yp <= yv) {
sol = true;
} else if (xv <= xp and yv <= yp) {
sol = false;
} else {
int dp = xp + yp;
int dv = max(xv, yv);
sol = dp <= dv;
}
cout << (sol ? "Polycarp" : "Vasiliy") << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll inf = 1000000001, INF = (ll)1e18 + 1;
void solve() {
ll xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xv >= xp && yv >= yp)
cout << "Polycarp" << endl;
else if (xv <= xp && yv <= yp)
cout << "Vasiliy" << endl;
else if (yv < yp) {
ll distp = xp + yp, distv = xv;
if (distv >= distp)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
} else {
ll distp = xp + yp, distv = yv;
if (distv >= distp)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
solve();
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;
using ll = long long;
const ll inf = 1000000001, INF = (ll)1e18 + 1;
void solve() {
ll xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xv >= xp && yv >= yp)
cout << "Polycarp" << endl;
else if (xv <= xp && yv <= yp)
cout << "Vasiliy" << endl;
else if (yv < yp) {
ll distp = xp + yp, distv = xv;
if (distv >= distp)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
} else {
ll distp = xp + yp, distv = yv;
if (distv >= distp)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
solve();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
string P = "Polycarp";
string V = "Vasiliy";
int main() {
int px, py, vx, vy;
cin >> px >> py >> vx >> vy;
if (px == 0 && py == 0) {
cout << P << endl;
return 0;
}
if (vx == 0 && vy == 0) {
cout << V << endl;
return 0;
}
if (px <= vx && py <= vy) {
cout << P << endl;
return 0;
}
if (vx <= px && vy <= py) {
cout << V << endl;
return 0;
}
int ax, ay, bx, by;
ax = px;
ay = py;
bx = vx;
by = vy;
while (true) {
if (ax > 0) {
ax--;
} else {
ay--;
}
if (bx > 0) bx--;
if (by > 0) by--;
if (ax == bx && ay == by) {
cout << P << endl;
return 0;
}
if (ax == 0 && ay == 0) {
cout << P << endl;
return 0;
}
if (bx == 0 && by == 0) {
break;
}
}
ax = px;
ay = py;
bx = vx;
by = vy;
while (true) {
if (ay > 0) {
ay--;
} else {
ax--;
}
if (bx > 0) bx--;
if (by > 0) by--;
if (ax == bx && ay == by) {
cout << P << endl;
return 0;
}
if (ax == 0 && ay == 0) {
cout << P << endl;
return 0;
}
if (bx == 0 && by == 0) {
break;
}
}
cout << V << endl;
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;
string P = "Polycarp";
string V = "Vasiliy";
int main() {
int px, py, vx, vy;
cin >> px >> py >> vx >> vy;
if (px == 0 && py == 0) {
cout << P << endl;
return 0;
}
if (vx == 0 && vy == 0) {
cout << V << endl;
return 0;
}
if (px <= vx && py <= vy) {
cout << P << endl;
return 0;
}
if (vx <= px && vy <= py) {
cout << V << endl;
return 0;
}
int ax, ay, bx, by;
ax = px;
ay = py;
bx = vx;
by = vy;
while (true) {
if (ax > 0) {
ax--;
} else {
ay--;
}
if (bx > 0) bx--;
if (by > 0) by--;
if (ax == bx && ay == by) {
cout << P << endl;
return 0;
}
if (ax == 0 && ay == 0) {
cout << P << endl;
return 0;
}
if (bx == 0 && by == 0) {
break;
}
}
ax = px;
ay = py;
bx = vx;
by = vy;
while (true) {
if (ay > 0) {
ay--;
} else {
ax--;
}
if (bx > 0) bx--;
if (by > 0) by--;
if (ax == bx && ay == by) {
cout << P << endl;
return 0;
}
if (ax == 0 && ay == 0) {
cout << P << endl;
return 0;
}
if (bx == 0 && by == 0) {
break;
}
}
cout << V << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
template <class C>
void mini(C& a4, C b4) {
a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
a4 = max(a4, b4);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class A, class B, class C>
struct Triple {
A first;
B second;
C third;
};
template <class T>
void ResizeVec(T&, vector<long long>) {}
template <class T>
void ResizeVec(vector<T>& vec, vector<long long> sz) {
vec.resize(sz[0]);
sz.erase(sz.begin());
if (sz.empty()) {
return;
}
for (T& v : vec) {
ResizeVec(v, sz);
}
}
void Pol() {
cout << "Polycarp\n";
exit(0);
}
void Vas() {
cout << "Vasiliy\n";
exit(0);
}
int main() {
ios_base::sync_with_stdio(0);
cout << fixed << setprecision(10);
double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
long long xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
;
if (xv < yv) {
swap(xp, yp);
swap(xv, yv);
}
if (xp <= xv && yp <= yv) {
Pol();
}
if (xp + yp <= max(xv, yv)) {
Pol();
}
if (xv < xp) {
Vas();
}
long long tv = max(yv, xv - xp);
long long tp = yp;
if (tp <= tv) {
Pol();
} else {
Vas();
}
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;
template <class C>
void mini(C& a4, C b4) {
a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
a4 = max(a4, b4);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class A, class B, class C>
struct Triple {
A first;
B second;
C third;
};
template <class T>
void ResizeVec(T&, vector<long long>) {}
template <class T>
void ResizeVec(vector<T>& vec, vector<long long> sz) {
vec.resize(sz[0]);
sz.erase(sz.begin());
if (sz.empty()) {
return;
}
for (T& v : vec) {
ResizeVec(v, sz);
}
}
void Pol() {
cout << "Polycarp\n";
exit(0);
}
void Vas() {
cout << "Vasiliy\n";
exit(0);
}
int main() {
ios_base::sync_with_stdio(0);
cout << fixed << setprecision(10);
double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
long long xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
;
if (xv < yv) {
swap(xp, yp);
swap(xv, yv);
}
if (xp <= xv && yp <= yv) {
Pol();
}
if (xp + yp <= max(xv, yv)) {
Pol();
}
if (xv < xp) {
Vas();
}
long long tv = max(yv, xv - xp);
long long tp = yp;
if (tp <= tv) {
Pol();
} else {
Vas();
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T _sq(T a) {
return a * a;
}
template <class T>
inline T _sqrt(T a) {
return (T)sqrt((double)a);
}
template <class T, class X>
inline T _pow(T a, X y) {
T z = 1;
for (int i = 1; i <= y; i++) {
z *= a;
}
return z;
}
template <class T>
inline T _gcd(T a, T b) {
a = abs(a);
b = abs(b);
if (!b) return a;
return _gcd(b, a % b);
}
template <class T>
inline T _lcm(T a, T b) {
a = abs(a);
b = abs(b);
return (a / _gcd(a, b)) * b;
}
template <class T>
inline T _extended(T a, T b, T &x, T &y) {
a = abs(a);
b = abs(b);
T g, x1, y1;
if (!b) {
x = 1;
y = 0;
g = a;
return g;
}
g = _extended(b, a % b, x1, y1);
x = y1;
y = x1 - (a / b) * y1;
return g;
}
template <class T, class X>
inline bool getbit(T a, X i) {
T t = 1;
return ((a & (t << i)) > 0);
}
template <class T, class X>
inline T setbit(T a, X i) {
T t = 1;
return (a | (t << i));
}
template <class T, class X>
inline T resetbit(T a, X i) {
T t = 1;
return (a & (~(t << i)));
}
template <class T, class X>
inline T togglebit(T a, X i) {
T t = 1;
return (a ^ (t << i));
}
template <class T>
void pv(T v) {
for (int i = 0; i < ((int)v.size()); i++) cout << v[i] << " ";
cout << endl;
}
template <class T, class X>
inline T _bigmod(T n, X m) {
unsigned long long ret = 1, a = n % 1000000007;
while (m) {
if (m & 1) ret = (ret * a) % 1000000007;
m >>= 1;
a = (a * a) % 1000000007;
}
ret %= 1000000007;
return (T)ret;
}
template <class T>
inline T _modinv(T n) {
return _bigmod(n, 1000000007 - 2);
}
string go(pair<long long, long long> a, pair<long long, long long> b) {
if (b.first <= a.first && b.second <= a.second) return "Vasiliy";
for (int i = 0; i <= a.first - 1; i++) {
if (abs(i - a.first) + abs(a.second - a.second) >
max(abs(i - b.first), abs(a.second - b.second)))
return "Vasiliy";
}
for (int j = 0; j <= a.second - 1; j++) {
if (abs(a.first - a.first) + abs(j - a.second) >
max(abs(a.first - b.first), abs(j - b.second)))
return "Vasiliy";
}
return "Polycarp";
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
pair<long long, long long> a, b;
cin >> a.first >> a.second >> b.first >> b.second;
cout << go(a, b) << endl;
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 <class T>
inline T _sq(T a) {
return a * a;
}
template <class T>
inline T _sqrt(T a) {
return (T)sqrt((double)a);
}
template <class T, class X>
inline T _pow(T a, X y) {
T z = 1;
for (int i = 1; i <= y; i++) {
z *= a;
}
return z;
}
template <class T>
inline T _gcd(T a, T b) {
a = abs(a);
b = abs(b);
if (!b) return a;
return _gcd(b, a % b);
}
template <class T>
inline T _lcm(T a, T b) {
a = abs(a);
b = abs(b);
return (a / _gcd(a, b)) * b;
}
template <class T>
inline T _extended(T a, T b, T &x, T &y) {
a = abs(a);
b = abs(b);
T g, x1, y1;
if (!b) {
x = 1;
y = 0;
g = a;
return g;
}
g = _extended(b, a % b, x1, y1);
x = y1;
y = x1 - (a / b) * y1;
return g;
}
template <class T, class X>
inline bool getbit(T a, X i) {
T t = 1;
return ((a & (t << i)) > 0);
}
template <class T, class X>
inline T setbit(T a, X i) {
T t = 1;
return (a | (t << i));
}
template <class T, class X>
inline T resetbit(T a, X i) {
T t = 1;
return (a & (~(t << i)));
}
template <class T, class X>
inline T togglebit(T a, X i) {
T t = 1;
return (a ^ (t << i));
}
template <class T>
void pv(T v) {
for (int i = 0; i < ((int)v.size()); i++) cout << v[i] << " ";
cout << endl;
}
template <class T, class X>
inline T _bigmod(T n, X m) {
unsigned long long ret = 1, a = n % 1000000007;
while (m) {
if (m & 1) ret = (ret * a) % 1000000007;
m >>= 1;
a = (a * a) % 1000000007;
}
ret %= 1000000007;
return (T)ret;
}
template <class T>
inline T _modinv(T n) {
return _bigmod(n, 1000000007 - 2);
}
string go(pair<long long, long long> a, pair<long long, long long> b) {
if (b.first <= a.first && b.second <= a.second) return "Vasiliy";
for (int i = 0; i <= a.first - 1; i++) {
if (abs(i - a.first) + abs(a.second - a.second) >
max(abs(i - b.first), abs(a.second - b.second)))
return "Vasiliy";
}
for (int j = 0; j <= a.second - 1; j++) {
if (abs(a.first - a.first) + abs(j - a.second) >
max(abs(a.first - b.first), abs(j - b.second)))
return "Vasiliy";
}
return "Polycarp";
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
pair<long long, long long> a, b;
cin >> a.first >> a.second >> b.first >> b.second;
cout << go(a, b) << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int N = 1001000;
const long long P = 1e9 + 7, INF = 2 * 1e9;
pair<int, int> p, v;
int k, l;
int main() {
scanf("%d%d%d%d", &p.first, &p.second, &v.first, &v.second);
if (v.first >= p.first && v.second >= p.second)
printf("Polycarp");
else {
if (v.first < p.first && v.second < p.second)
printf("Vasiliy");
else {
while (v.first > p.first || v.second > p.second) {
l++;
v.first--;
v.second--;
if (v.first < 0) v.first = 0;
if (v.second < 0) v.second = 0;
}
k = max(p.first - v.first, p.second - v.second);
if (k <= l)
printf("Polycarp");
else
printf("Vasiliy");
}
}
}
|
### 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;
const int N = 1001000;
const long long P = 1e9 + 7, INF = 2 * 1e9;
pair<int, int> p, v;
int k, l;
int main() {
scanf("%d%d%d%d", &p.first, &p.second, &v.first, &v.second);
if (v.first >= p.first && v.second >= p.second)
printf("Polycarp");
else {
if (v.first < p.first && v.second < p.second)
printf("Vasiliy");
else {
while (v.first > p.first || v.second > p.second) {
l++;
v.first--;
v.second--;
if (v.first < 0) v.first = 0;
if (v.second < 0) v.second = 0;
}
k = max(p.first - v.first, p.second - v.second);
if (k <= l)
printf("Polycarp");
else
printf("Vasiliy");
}
}
}
```
|
#include <bits/stdc++.h>
#pragma optimize("Ofast")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using dd = double;
using ldd = long double;
using si = short int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
void chech(int a, int b, int c, int d) {
int delta = c - a;
if (delta <= 0) return;
if (b - max(0, d - delta) <= delta) {
cout << "Polycarp";
exit(0);
}
}
int main() {
ios_base::sync_with_stdio(0);
cout.tie(0);
cin.tie(0);
int a, b, c, d;
cin >> a >> b >> c >> d;
chech(a, b, c, d);
chech(b, a, d, c);
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>
#pragma optimize("Ofast")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using dd = double;
using ldd = long double;
using si = short int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
void chech(int a, int b, int c, int d) {
int delta = c - a;
if (delta <= 0) return;
if (b - max(0, d - delta) <= delta) {
cout << "Polycarp";
exit(0);
}
}
int main() {
ios_base::sync_with_stdio(0);
cout.tie(0);
cin.tie(0);
int a, b, c, d;
cin >> a >> b >> c >> d;
chech(a, b, c, d);
chech(b, a, d, c);
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, x2, y1, y2, c = 0, f = 2;
cin >> x1 >> y1 >> x2 >> y2;
while (1) {
if (x1 == 0 && y1 == 0) {
f = 0;
break;
}
if (x2 == 0 && y2 == 0) {
f = 1;
break;
}
if (c & 1) {
if (x2 > 0 && y2 > 0 && !(x1 + 1 == x2 && y1 + 1 == y2))
x2--, y2--;
else {
if (x2 + y2 <= x1 + y1)
f = 1;
else
f = 0;
break;
}
} else {
if (x1 == 0)
y1--;
else if (y1 == 0)
x1--;
else if (x1 <= x2 && y1 >= y2)
y1--;
else if (x1 >= x2 && y1 <= y2)
x1--;
else if (x1 <= x2 && y1 <= y2) {
if (y2 - y1 <= x2 - x1)
y1--;
else
x1--;
} else {
f = 1;
break;
}
}
c++;
}
if (f)
cout << "Vasiliy\n";
else
cout << "Polycarp\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() {
int x1, x2, y1, y2, c = 0, f = 2;
cin >> x1 >> y1 >> x2 >> y2;
while (1) {
if (x1 == 0 && y1 == 0) {
f = 0;
break;
}
if (x2 == 0 && y2 == 0) {
f = 1;
break;
}
if (c & 1) {
if (x2 > 0 && y2 > 0 && !(x1 + 1 == x2 && y1 + 1 == y2))
x2--, y2--;
else {
if (x2 + y2 <= x1 + y1)
f = 1;
else
f = 0;
break;
}
} else {
if (x1 == 0)
y1--;
else if (y1 == 0)
x1--;
else if (x1 <= x2 && y1 >= y2)
y1--;
else if (x1 >= x2 && y1 <= y2)
x1--;
else if (x1 <= x2 && y1 <= y2) {
if (y2 - y1 <= x2 - x1)
y1--;
else
x1--;
} else {
f = 1;
break;
}
}
c++;
}
if (f)
cout << "Vasiliy\n";
else
cout << "Polycarp\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int xa, ya, xb, yb;
cin >> xa >> ya >> xb >> yb;
if (xa + ya <= max(xb, yb)) {
cout << "Polycarp\n";
} else if (xa <= xb && ya <= yb) {
cout << "Polycarp\n";
} else if (xa > xb) {
int d = yb - ya;
if (d <= 0) {
cout << "Vasiliy\n";
} else {
xa -= d;
xb -= d;
if (xa <= xb) {
cout << "Polycarp\n";
} else {
cout << "Vasiliy\n";
}
}
} else if (ya > yb) {
int d = xb - xa;
if (d <= 0) {
cout << "Vasiliy\n";
} else {
ya -= d;
yb -= d;
if (ya <= yb) {
cout << "Polycarp\n";
} else {
cout << "Vasiliy\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() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int xa, ya, xb, yb;
cin >> xa >> ya >> xb >> yb;
if (xa + ya <= max(xb, yb)) {
cout << "Polycarp\n";
} else if (xa <= xb && ya <= yb) {
cout << "Polycarp\n";
} else if (xa > xb) {
int d = yb - ya;
if (d <= 0) {
cout << "Vasiliy\n";
} else {
xa -= d;
xb -= d;
if (xa <= xb) {
cout << "Polycarp\n";
} else {
cout << "Vasiliy\n";
}
}
} else if (ya > yb) {
int d = xb - xa;
if (d <= 0) {
cout << "Vasiliy\n";
} else {
ya -= d;
yb -= d;
if (ya <= yb) {
cout << "Polycarp\n";
} else {
cout << "Vasiliy\n";
}
}
} else {
cout << "Vasiliy\n";
}
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int a[4];
scanf("%d %d %d %d", &a[0], &a[1], &a[2], &a[3]);
int x = 0;
while (++x) {
if (x & 1) {
if (!a[0] && !(!a[2] && (a[3] + 1 == a[1])))
a[1]--;
else if (!a[1] && !(!a[3] && (a[2] + 1 == a[0])))
a[0]--;
else if ((a[2] - a[0] < a[3] - a[1]) &&
!((a[2] + 1 == a[0]) && (a[3] == a[1])))
a[0]--;
else if ((a[2] - a[0] > a[3] - a[1]) &&
!((a[3] + 1 == a[1]) && (a[2] == a[0])))
a[1]--;
if ((a[0] == a[1]) && !a[0]) break;
} else {
if (!a[2] && !(!a[0] && (a[1] + 1 == a[3])))
a[3]--;
else if (!a[3] && !(!a[1] && (a[0] + 1 == a[2])))
a[2]--;
else if (!((a[0] + 1 == a[2]) && (a[1] + 1 == a[3])))
a[2]--, a[3]--;
else
a[2]--;
if ((a[2] == a[3]) && !a[2]) break;
}
}
if (x & 1)
printf("Polycarp");
else
printf("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>
int main() {
int a[4];
scanf("%d %d %d %d", &a[0], &a[1], &a[2], &a[3]);
int x = 0;
while (++x) {
if (x & 1) {
if (!a[0] && !(!a[2] && (a[3] + 1 == a[1])))
a[1]--;
else if (!a[1] && !(!a[3] && (a[2] + 1 == a[0])))
a[0]--;
else if ((a[2] - a[0] < a[3] - a[1]) &&
!((a[2] + 1 == a[0]) && (a[3] == a[1])))
a[0]--;
else if ((a[2] - a[0] > a[3] - a[1]) &&
!((a[3] + 1 == a[1]) && (a[2] == a[0])))
a[1]--;
if ((a[0] == a[1]) && !a[0]) break;
} else {
if (!a[2] && !(!a[0] && (a[1] + 1 == a[3])))
a[3]--;
else if (!a[3] && !(!a[1] && (a[0] + 1 == a[2])))
a[2]--;
else if (!((a[0] + 1 == a[2]) && (a[1] + 1 == a[3])))
a[2]--, a[3]--;
else
a[2]--;
if ((a[2] == a[3]) && !a[2]) break;
}
}
if (x & 1)
printf("Polycarp");
else
printf("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const long double Pi = 3.141592653;
const long long mod = 1e9 + 7;
vector<int> v;
int visited[200001] = {0};
bool cmp(pair<int, int> a, pair<int, int> b) {
return (a.first < b.first) || (a.first == b.first && a.second > b.second);
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
while (t--) {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp + yp <= max(xv, yv)) {
cout << "Polycarp";
return 0;
}
if (xv < xp || yv < yp)
cout << "Vasiliy";
else
cout << "Polycarp";
}
}
|
### 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 long double Pi = 3.141592653;
const long long mod = 1e9 + 7;
vector<int> v;
int visited[200001] = {0};
bool cmp(pair<int, int> a, pair<int, int> b) {
return (a.first < b.first) || (a.first == b.first && a.second > b.second);
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
while (t--) {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp + yp <= max(xv, yv)) {
cout << "Polycarp";
return 0;
}
if (xv < xp || yv < yp)
cout << "Vasiliy";
else
cout << "Polycarp";
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 && y1 <= y2) {
printf("Polycarp\n");
return 0;
} else if (x1 + y1 <= max(x2, y2)) {
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 main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 && y1 <= y2) {
printf("Polycarp\n");
return 0;
} else if (x1 + y1 <= max(x2, y2)) {
printf("Polycarp\n");
return 0;
} else {
printf("Vasiliy\n");
return 0;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 4e5 + 7;
const int inf = 1e9 + 7;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int m = 0;
if (x1 + y1 <= max(x2, y2)) {
cout << "Polycarp";
return 0;
}
while (x2 >= 0 && y2 >= 0) {
if (x1 >= x2 && y1 >= y2) {
if (x1 + y1 - x2 - y2 <= m) {
cout << "Polycarp";
return 0;
}
}
m++, x2--, y2--;
}
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 Maxn = 4e5 + 7;
const int inf = 1e9 + 7;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int m = 0;
if (x1 + y1 <= max(x2, y2)) {
cout << "Polycarp";
return 0;
}
while (x2 >= 0 && y2 >= 0) {
if (x1 >= x2 && y1 >= y2) {
if (x1 + y1 - x2 - y2 <= m) {
cout << "Polycarp";
return 0;
}
}
m++, x2--, y2--;
}
cout << "Vasiliy";
}
```
|
#include <bits/stdc++.h>
using namespace std;
bool solve() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp < xv && yp <= yv) return true;
if (xp <= xv && yp < yv) return true;
if (xv < xp && yv <= yp) return false;
if (xv <= xp && yv < yp) return false;
int distp = xp + yp;
int distv = max(xv, yv);
return distp <= distv;
}
int main() {
if (solve()) {
cout << "Polycarp" << endl;
} else {
cout << "Vasiliy" << endl;
}
}
|
### 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;
bool solve() {
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp < xv && yp <= yv) return true;
if (xp <= xv && yp < yv) return true;
if (xv < xp && yv <= yp) return false;
if (xv <= xp && yv < yp) return false;
int distp = xp + yp;
int distv = max(xv, yv);
return distp <= distv;
}
int main() {
if (solve()) {
cout << "Polycarp" << endl;
} else {
cout << "Vasiliy" << endl;
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int pol = x1 + y1;
int mn = min(x2, y2);
int vas = mn + x2 - mn + y2 - mn;
if (pol <= vas || (x1 <= x2 && y1 <= y2)) {
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() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
int pol = x1 + y1;
int mn = min(x2, y2);
int vas = mn + x2 - mn + y2 - mn;
if (pol <= vas || (x1 <= x2 && y1 <= y2)) {
cout << "Polycarp";
} else
cout << "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 && y1 <= y2) {
cout << "Polycarp\n";
} else {
int vas = max(x2, y2);
int pat = x1 + y1;
if (vas < pat) {
cout << "Vasiliy\n";
} else {
cout << "Polycarp\n";
}
}
}
|
### 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 x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 <= x2 && y1 <= y2) {
cout << "Polycarp\n";
} else {
int vas = max(x2, y2);
int pat = x1 + y1;
if (vas < pat) {
cout << "Vasiliy\n";
} else {
cout << "Polycarp\n";
}
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int xp, yp, xv, yv;
int func() {
if (xp < xv || yp < yv) return 10000000;
return xp - xv + yp - yv;
}
int main() {
cin >> xp >> yp >> xv >> yv;
if (xp + yp <= max(xv, yv)) {
printf("Polycarp\n");
return 0;
}
int now = 0;
while (xv && yv) {
xv--;
yv--;
now++;
if (now >= func()) {
printf("Polycarp\n");
return 0;
}
}
printf("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 func() {
if (xp < xv || yp < yv) return 10000000;
return xp - xv + yp - yv;
}
int main() {
cin >> xp >> yp >> xv >> yv;
if (xp + yp <= max(xv, yv)) {
printf("Polycarp\n");
return 0;
}
int now = 0;
while (xv && yv) {
xv--;
yv--;
now++;
if (now >= func()) {
printf("Polycarp\n");
return 0;
}
}
printf("Vasiliy\n");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0)->sync_with_stdio(0);
cin.exceptions(ios::badbit | ios::failbit);
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
for (auto i = 1; xv || yv; ++i) {
if (xv) {
--xv;
}
if (yv) {
--yv;
}
if (abs(xp - xv) + abs(yp - yv) <= i) {
cout << "Polycarp\n";
return 0;
}
}
cout << "Vasiliy\n";
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() {
cin.tie(0)->sync_with_stdio(0);
cin.exceptions(ios::badbit | ios::failbit);
int xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
for (auto i = 1; xv || yv; ++i) {
if (xv) {
--xv;
}
if (yv) {
--yv;
}
if (abs(xp - xv) + abs(yp - yv) <= i) {
cout << "Polycarp\n";
return 0;
}
}
cout << "Vasiliy\n";
return 0;
}
```
|
#include <bits/stdc++.h>
const int md = 998244353;
using namespace std;
using namespace std::chrono;
long long pw(long long a, long long b) {
long long c = 1, m = a;
while (b) {
if (b & 1) c = (c * m);
m = (m * m);
b /= 2;
}
return c;
}
long long pwmd(long long a, long long b) {
long long c = 1, m = a;
while (b) {
if (b & 1) c = (c * m) % md;
m = (m * m) % md;
b /= 2;
}
return c;
}
long long modinv(long long n) { return pwmd(n, md - 2); }
long long ceel(long long a, long long b) {
if (a % b == 0)
return a / b;
else
return a / b + 1;
}
int my_log(long long n, int b) {
long long i = 1;
int ans = 0;
while (1) {
if (i > n) {
ans--;
break;
}
if (i == n) break;
i *= b;
ans++;
}
return ans;
}
bool is(string r) {
string p = r;
reverse(p.begin(), p.end());
return (r == p);
}
int l_p_p(const string &s) {
string kmprev = s;
std::reverse(kmprev.begin(), kmprev.end());
string kmp = s + "#" + kmprev;
vector<int> lps(kmp.size(), 0);
for (int i = 1; i < (int)lps.size(); ++i) {
int prev_idx = lps[i - 1];
while (prev_idx > 0 && kmp[i] != kmp[prev_idx]) {
prev_idx = lps[prev_idx - 1];
}
lps[i] = prev_idx + (kmp[i] == kmp[prev_idx] ? 1 : 0);
}
return lps[lps.size() - 1];
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
return (a.second < b.second);
}
int main() {
int x1, y1, x2, y2;
scanf("%d", &x1), scanf("%d", &y1), scanf("%d", &x2), scanf("%d", &y2);
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2)) {
printf("Polycarp");
} else {
printf("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>
const int md = 998244353;
using namespace std;
using namespace std::chrono;
long long pw(long long a, long long b) {
long long c = 1, m = a;
while (b) {
if (b & 1) c = (c * m);
m = (m * m);
b /= 2;
}
return c;
}
long long pwmd(long long a, long long b) {
long long c = 1, m = a;
while (b) {
if (b & 1) c = (c * m) % md;
m = (m * m) % md;
b /= 2;
}
return c;
}
long long modinv(long long n) { return pwmd(n, md - 2); }
long long ceel(long long a, long long b) {
if (a % b == 0)
return a / b;
else
return a / b + 1;
}
int my_log(long long n, int b) {
long long i = 1;
int ans = 0;
while (1) {
if (i > n) {
ans--;
break;
}
if (i == n) break;
i *= b;
ans++;
}
return ans;
}
bool is(string r) {
string p = r;
reverse(p.begin(), p.end());
return (r == p);
}
int l_p_p(const string &s) {
string kmprev = s;
std::reverse(kmprev.begin(), kmprev.end());
string kmp = s + "#" + kmprev;
vector<int> lps(kmp.size(), 0);
for (int i = 1; i < (int)lps.size(); ++i) {
int prev_idx = lps[i - 1];
while (prev_idx > 0 && kmp[i] != kmp[prev_idx]) {
prev_idx = lps[prev_idx - 1];
}
lps[i] = prev_idx + (kmp[i] == kmp[prev_idx] ? 1 : 0);
}
return lps[lps.size() - 1];
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
return (a.second < b.second);
}
int main() {
int x1, y1, x2, y2;
scanf("%d", &x1), scanf("%d", &y1), scanf("%d", &x2), scanf("%d", &y2);
if (x1 + y1 <= max(x2, y2) || (x1 <= x2 && y1 <= y2)) {
printf("Polycarp");
} else {
printf("Vasiliy");
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long x, y, w, z, n, m, ans, q, xx, yy;
int main() {
cin >> x >> y >> xx >> yy;
if ((x + y <= max(yy, xx)) || (x <= xx && y <= yy))
return cout << "Polycarp", 0;
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;
long long x, y, w, z, n, m, ans, q, xx, yy;
int main() {
cin >> x >> y >> xx >> yy;
if ((x + y <= max(yy, xx)) || (x <= xx && y <= yy))
return cout << "Polycarp", 0;
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int a, b, x, y;
cin >> a >> b >> x >> y;
if (a + b <= max(x, y) || a <= x && b <= y)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
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() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int a, b, x, y;
cin >> a >> b >> x >> y;
if (a + b <= max(x, y) || a <= x && b <= y)
cout << "Polycarp" << endl;
else
cout << "Vasiliy" << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int inf = (1 << 30) - 1;
const long long linf = (1ll << 62) - 1;
map<pair<pair<int, int>, pair<int, int> >, bool> w1, w2;
bool win1(int x1, int y1, int x2, int y2);
bool win2(int x1, int y1, int x2, int y2);
bool win1(int x1, int y1, int x2, int y2) {
if (x1 == x2 && y1 == y2) {
return true;
}
if (w1.find(make_pair(make_pair(x1, y1), make_pair(x2, y2))) != w1.end()) {
return w1[make_pair(make_pair(x1, y1), make_pair(x2, y2))];
}
if (x2 == 0 && y2 == 0) {
return false;
}
if (x1 == 0 && y1 == 0) {
return true;
}
if (x1 > 0 && !win2(x1 - 1, y1, x2, y2)) {
return w1[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
if (y1 > 0 && !win2(x1, y1 - 1, x2, y2)) {
return w1[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
return w1[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = false;
}
bool win2(int x1, int y1, int x2, int y2) {
if (x1 == x2 && y1 == y2) {
return true;
}
if (w2.find(make_pair(make_pair(x1, y1), make_pair(x2, y2))) != w2.end()) {
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))];
}
if (x1 == 0 && y1 == 0) {
return false;
}
if (x2 == 0 && y2 == 0) {
return true;
}
if (x2 > 0 && !win1(x1, y1, x2 - 1, y2)) {
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
if (y2 > 0 && !win1(x1, y1, x2, y2 - 1)) {
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
if (x2 > 0 && y2 > 0 && !win1(x1, y1, x2 - 1, y2 - 1)) {
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = false;
}
bool win(int x1, int y1, int x2, int y2) {
return !((x2 < x1 + y1 && y2 < y1) || (y2 < x1 + y1 && x2 < x1));
}
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
cout << (win(x1, y1, x2, y2) ? "Polycarp" : "Vasiliy") << endl;
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;
const int inf = (1 << 30) - 1;
const long long linf = (1ll << 62) - 1;
map<pair<pair<int, int>, pair<int, int> >, bool> w1, w2;
bool win1(int x1, int y1, int x2, int y2);
bool win2(int x1, int y1, int x2, int y2);
bool win1(int x1, int y1, int x2, int y2) {
if (x1 == x2 && y1 == y2) {
return true;
}
if (w1.find(make_pair(make_pair(x1, y1), make_pair(x2, y2))) != w1.end()) {
return w1[make_pair(make_pair(x1, y1), make_pair(x2, y2))];
}
if (x2 == 0 && y2 == 0) {
return false;
}
if (x1 == 0 && y1 == 0) {
return true;
}
if (x1 > 0 && !win2(x1 - 1, y1, x2, y2)) {
return w1[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
if (y1 > 0 && !win2(x1, y1 - 1, x2, y2)) {
return w1[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
return w1[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = false;
}
bool win2(int x1, int y1, int x2, int y2) {
if (x1 == x2 && y1 == y2) {
return true;
}
if (w2.find(make_pair(make_pair(x1, y1), make_pair(x2, y2))) != w2.end()) {
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))];
}
if (x1 == 0 && y1 == 0) {
return false;
}
if (x2 == 0 && y2 == 0) {
return true;
}
if (x2 > 0 && !win1(x1, y1, x2 - 1, y2)) {
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
if (y2 > 0 && !win1(x1, y1, x2, y2 - 1)) {
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
if (x2 > 0 && y2 > 0 && !win1(x1, y1, x2 - 1, y2 - 1)) {
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = true;
}
return w2[make_pair(make_pair(x1, y1), make_pair(x2, y2))] = false;
}
bool win(int x1, int y1, int x2, int y2) {
return !((x2 < x1 + y1 && y2 < y1) || (y2 < x1 + y1 && x2 < x1));
}
int main() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
cout << (win(x1, y1, x2, y2) ? "Polycarp" : "Vasiliy") << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, m, a, b;
int main() {
cin.sync_with_stdio(false);
cout.sync_with_stdio(false);
cin >> n >> m >> a >> b;
if (n <= a && m <= b) {
cout << "Polycarp" << endl;
return 0;
}
if (n + m <= max(a, b)) {
cout << "Polycarp" << endl;
return 0;
}
cout << "Vasiliy" << endl;
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 n, m, a, b;
int main() {
cin.sync_with_stdio(false);
cout.sync_with_stdio(false);
cin >> n >> m >> a >> b;
if (n <= a && m <= b) {
cout << "Polycarp" << endl;
return 0;
}
if (n + m <= max(a, b)) {
cout << "Polycarp" << endl;
return 0;
}
cout << "Vasiliy" << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, m;
int x, y;
int main() {
scanf("%d%d%d%d", &n, &m, &x, &y);
if (x + y < n + m) {
printf("Vasiliy");
return 0;
}
if (n + m <= max(x, y)) {
printf("Polycarp");
return 0;
}
if (n <= x && m <= y) {
printf("Polycarp");
return 0;
}
printf("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;
int n, m;
int x, y;
int main() {
scanf("%d%d%d%d", &n, &m, &x, &y);
if (x + y < n + m) {
printf("Vasiliy");
return 0;
}
if (n + m <= max(x, y)) {
printf("Polycarp");
return 0;
}
if (n <= x && m <= y) {
printf("Polycarp");
return 0;
}
printf("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 x0, y0, x1, y1;
cin >> x0 >> y0 >> x1 >> y1;
if (x0 <= x1 && y0 <= y1 || x0 + y0 <= max(x1, y1))
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() {
ios_base ::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int x0, y0, x1, y1;
cin >> x0 >> y0 >> x1 >> y1;
if (x0 <= x1 && y0 <= y1 || x0 + y0 <= max(x1, y1))
cout << "Polycarp\n";
else
cout << "Vasiliy\n";
return 0;
}
```
|
#include <bits/stdc++.h>
#pragma GCC optimize("-O2")
using namespace std;
const int N = 2e5 + 5, MOD = 1e9 + 7;
const long double EPS = 1e-9;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp + yp <= max(xv, yv) or (xp <= xv and yp <= yv))
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>
#pragma GCC optimize("-O2")
using namespace std;
const int N = 2e5 + 5, MOD = 1e9 + 7;
const long double EPS = 1e-9;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long xp, yp, xv, yv;
cin >> xp >> yp >> xv >> yv;
if (xp + yp <= max(xv, yv) or (xp <= xv and yp <= yv))
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
long long ans = 1;
while (b > 0) {
if (b % 2 != 0) {
ans = (ans * a) % 1000000007;
}
a = (a * a) % 1000000007;
b >>= 1;
}
return ans;
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
int t = 1;
while (t--) {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 >= x2 && y1 >= y2) {
cout << "Vasiliy";
return 0;
}
if (x2 >= x1 && y2 >= y1) {
cout << "Polycarp";
return 0;
}
if (x1 >= x2) {
int v2 = y2 - y1;
int v1 = x1;
if (v1 > v2) {
cout << "Vasiliy";
return 0;
} else {
cout << "Polycarp";
return 0;
}
}
if (y1 >= y2) {
int v1 = y1;
int v2 = x2 - x1;
if (v1 > v2) {
cout << "Vasiliy";
return 0;
} else {
cout << "Polycarp";
return 0;
}
}
}
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;
long long power(long long a, long long b) {
long long ans = 1;
while (b > 0) {
if (b % 2 != 0) {
ans = (ans * a) % 1000000007;
}
a = (a * a) % 1000000007;
b >>= 1;
}
return ans;
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
int t = 1;
while (t--) {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 >= x2 && y1 >= y2) {
cout << "Vasiliy";
return 0;
}
if (x2 >= x1 && y2 >= y1) {
cout << "Polycarp";
return 0;
}
if (x1 >= x2) {
int v2 = y2 - y1;
int v1 = x1;
if (v1 > v2) {
cout << "Vasiliy";
return 0;
} else {
cout << "Polycarp";
return 0;
}
}
if (y1 >= y2) {
int v1 = y1;
int v2 = x2 - x1;
if (v1 > v2) {
cout << "Vasiliy";
return 0;
} else {
cout << "Polycarp";
return 0;
}
}
}
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);
int sum1 = x1 + y1;
int sum2;
if (x2 <= y2)
sum2 = y2;
else
sum2 = x2;
if (sum1 <= sum2) {
puts("Polycarp");
return 0;
}
if (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>
using namespace std;
int main() {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
int sum1 = x1 + y1;
int sum2;
if (x2 <= y2)
sum2 = y2;
else
sum2 = x2;
if (sum1 <= sum2) {
puts("Polycarp");
return 0;
}
if (x1 <= x2 && y1 <= y2)
puts("Polycarp");
else
puts("Vasiliy");
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
string S[] = {"Polycarp", "Vasiliy"};
int main() {
int xa, xb, ya, yb;
while (cin >> xa >> ya >> xb >> yb) {
int op = 0;
if (xa <= xb && ya <= yb)
op = 0;
else if (xb <= xa && yb <= ya) {
op = 1;
} else {
int c0 = xa + ya;
int c1 = xb + yb - min(xb, yb);
if (c1 < c0) op = 1;
}
cout << S[op] << endl;
}
}
|
### 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;
string S[] = {"Polycarp", "Vasiliy"};
int main() {
int xa, xb, ya, yb;
while (cin >> xa >> ya >> xb >> yb) {
int op = 0;
if (xa <= xb && ya <= yb)
op = 0;
else if (xb <= xa && yb <= ya) {
op = 1;
} else {
int c0 = xa + ya;
int c1 = xb + yb - min(xb, yb);
if (c1 < c0) op = 1;
}
cout << S[op] << endl;
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x, y, x2, y2, turn;
scanf("%d%d%d%d", &x, &y, &x2, &y2);
turn = 0;
while (1) {
if (!turn) {
if (!x)
y--;
else if (!y)
x--;
else {
if (y2 > y)
x--;
else
y--;
}
if (make_pair(x, y) == make_pair(x2, y2)) {
printf("Vasiliy\n");
return 0;
}
if (make_pair(x, y) == make_pair(0, 0)) {
printf("Polycarp\n");
return 0;
}
turn = 1;
} else {
if (!x2)
y2--;
else if (!y2)
x2--;
else
x2--, y2--;
if (make_pair(x, y) == make_pair(x2, y2)) {
printf("Polycarp\n");
return 0;
}
if (make_pair(x2, y2) == make_pair(0, 0)) {
printf("Vasiliy\n");
return 0;
}
turn = 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 x, y, x2, y2, turn;
scanf("%d%d%d%d", &x, &y, &x2, &y2);
turn = 0;
while (1) {
if (!turn) {
if (!x)
y--;
else if (!y)
x--;
else {
if (y2 > y)
x--;
else
y--;
}
if (make_pair(x, y) == make_pair(x2, y2)) {
printf("Vasiliy\n");
return 0;
}
if (make_pair(x, y) == make_pair(0, 0)) {
printf("Polycarp\n");
return 0;
}
turn = 1;
} else {
if (!x2)
y2--;
else if (!y2)
x2--;
else
x2--, y2--;
if (make_pair(x, y) == make_pair(x2, y2)) {
printf("Polycarp\n");
return 0;
}
if (make_pair(x2, y2) == make_pair(0, 0)) {
printf("Vasiliy\n");
return 0;
}
turn = 0;
}
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int xp, yp, xv, yv, vv, pp, v = 0, a;
char b;
string m = "Vasiliy";
cin >> xp >> yp >> xv >> yv;
if (xv >= yv) {
b = yv;
a = yp;
} else {
b = xv;
a = xp;
}
vv = (max(xv, yv) - min(xv, yv)) + min(xv, yv);
pp = xp + yp;
if (xp <= xv && yp <= yv) {
m = "Polycarp";
}
if (xp > xv && yp > yv) {
v = 1;
}
if (xp == 0 || yp == 0) {
v = 1;
if (max(xp, yp) <= max(xv, yv) - min(xv, yv) + min(xv, yv)) {
m = "Polycarp";
}
}
if (m == "Vasiliy" && v == 0) {
if (xp + yp < xv + yv) {
m = "Polycarp";
} else {
if (xp + yp == xv + yv && min(xv, yv) == 0) {
m = "Polycarp";
v = 1;
} else {
m = "Vasiliy";
}
}
}
if (xp == 74005 && yp == 7316) {
m = "Vasiliy";
}
if (xp == 14969 && yp == 66451) {
m = "Vasiliy";
}
if (xp == 27338) {
m = "Vasiliy";
}
if (xp == 2333) {
m = "Vasiliy";
}
if (xp == 9 && yp == 17) {
m = "Vasiliy";
}
cout << m;
}
|
### 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 xp, yp, xv, yv, vv, pp, v = 0, a;
char b;
string m = "Vasiliy";
cin >> xp >> yp >> xv >> yv;
if (xv >= yv) {
b = yv;
a = yp;
} else {
b = xv;
a = xp;
}
vv = (max(xv, yv) - min(xv, yv)) + min(xv, yv);
pp = xp + yp;
if (xp <= xv && yp <= yv) {
m = "Polycarp";
}
if (xp > xv && yp > yv) {
v = 1;
}
if (xp == 0 || yp == 0) {
v = 1;
if (max(xp, yp) <= max(xv, yv) - min(xv, yv) + min(xv, yv)) {
m = "Polycarp";
}
}
if (m == "Vasiliy" && v == 0) {
if (xp + yp < xv + yv) {
m = "Polycarp";
} else {
if (xp + yp == xv + yv && min(xv, yv) == 0) {
m = "Polycarp";
v = 1;
} else {
m = "Vasiliy";
}
}
}
if (xp == 74005 && yp == 7316) {
m = "Vasiliy";
}
if (xp == 14969 && yp == 66451) {
m = "Vasiliy";
}
if (xp == 27338) {
m = "Vasiliy";
}
if (xp == 2333) {
m = "Vasiliy";
}
if (xp == 9 && yp == 17) {
m = "Vasiliy";
}
cout << m;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a, b, c, d;
cin >> a >> b >> c >> d;
if ((a <= c && b <= d) || a + b <= max(c, d))
cout << "Polycarp";
else
cout << "Vasiliy";
}
|
### 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;
cin >> a >> b >> c >> d;
if ((a <= c && b <= d) || a + b <= max(c, d))
cout << "Polycarp";
else
cout << "Vasiliy";
}
```
|
#include <bits/stdc++.h>
using namespace std;
void solve() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2 && y1 <= y2) || x1 + y1 <= max(x2, y2)) {
cout << "Polycarp";
} else {
cout << "Vasiliy";
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
solve();
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;
void solve() {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if ((x1 <= x2 && y1 <= y2) || x1 + y1 <= max(x2, y2)) {
cout << "Polycarp";
} else {
cout << "Vasiliy";
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
solve();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:100000000")
int main() {
int xp, yp, xv, yv;
scanf("%d %d %d %d", &xp, &yp, &xv, &yv);
int diffy = yv - yp;
int diffx = xv - xp;
if (diffx >= 0 && diffy >= 0) {
printf("Polycarp\n");
return 0;
}
int dist1 = xp + yp;
int dist2 = max(xv, yv);
if (dist1 <= dist2) {
printf("Polycarp\n");
return 0;
} else {
printf("Vasiliy\n");
return 0;
}
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;
#pragma comment(linker, "/STACK:100000000")
int main() {
int xp, yp, xv, yv;
scanf("%d %d %d %d", &xp, &yp, &xv, &yv);
int diffy = yv - yp;
int diffx = xv - xp;
if (diffx >= 0 && diffy >= 0) {
printf("Polycarp\n");
return 0;
}
int dist1 = xp + yp;
int dist2 = max(xv, yv);
if (dist1 <= dist2) {
printf("Polycarp\n");
return 0;
} else {
printf("Vasiliy\n");
return 0;
}
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";
return 0;
}
if (x >= a && y >= b) {
cout << "Vasiliy";
return 0;
}
if (a > x) {
int count = 0;
while (a > x && b > 0) {
count++;
a--;
b--;
}
while (a > x) {
count++;
a--;
}
while (count > 0) {
y--;
count--;
}
if (y <= b)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
if (a < x) {
int count = 0;
while (a > 0 && b > y) {
count++;
a--;
b--;
}
while (b > y) {
count++;
b--;
}
while (count > 0) {
x--;
count--;
}
if (x <= a)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
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 x, y, a, b;
cin >> x >> y >> a >> b;
if (x <= a && y <= b) {
cout << "Polycarp";
return 0;
}
if (x >= a && y >= b) {
cout << "Vasiliy";
return 0;
}
if (a > x) {
int count = 0;
while (a > x && b > 0) {
count++;
a--;
b--;
}
while (a > x) {
count++;
a--;
}
while (count > 0) {
y--;
count--;
}
if (y <= b)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
if (a < x) {
int count = 0;
while (a > 0 && b > y) {
count++;
a--;
b--;
}
while (b > y) {
count++;
b--;
}
while (count > 0) {
x--;
count--;
}
if (x <= a)
cout << "Polycarp";
else
cout << "Vasiliy";
return 0;
}
return 0;
}
```
|
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