CodeForces 69D Dot （博弈+记忆化）

Description

Anton and Dasha like to play different games during breaks on checkered paper. By the 11th grade they managed to play all the games of this type and asked Vova the programmer to come up with a new game. Vova suggested to them to play a game under the code name "dot" with the following rules:

• On the checkered paper a coordinate system is drawn. A dot is initially put in the position (x, y).
• A move is shifting a dot to one of the pre-selected vectors. Also each player can once per game symmetrically reflect a dot relatively to the line y = x.
• Anton and Dasha take turns. Anton goes first.
• The player after whose move the distance from the dot to the coordinates' origin exceeds d, loses.

Help them to determine the winner.

Input

The first line of the input file contains 4 integers x, y, n, d ( - 200 ≤ x, y ≤ 200, 1 ≤ d ≤ 200, 1 ≤ n ≤ 20) — the initial coordinates of the dot, the distance d and the number of vectors. It is guaranteed that the initial dot is at the distance less than d from the origin of the coordinates. The following n lines each contain two non-negative numbers x_i and y_i (0 ≤ x_i, y_i ≤ 200) — the coordinates of the i-th vector. It is guaranteed that all the vectors are nonzero and different.

Output

You should print "Anton", if the winner is Anton in case of both players play the game optimally, and "Dasha" otherwise.

Sample Input

Input

0 0 2 3
1 1
1 2

Output

Anton

Input

0 0 2 4
1 1
1 2

Output

Dasha

题意：有一个移点的游戏，Anton先移，有n个移动选择，也可以沿着直线y=x对称且只能一次，如果有人先移动到距离原点>=d的时候为输

思路：对于直线y=x对称的情况，没有考虑，因为如果有人下一步一定移到>=d的位置的话，那对称是解决不了问题的，所以我们不考虑，现在设dfs(x, y)表示当前移动人是否能赢，一旦有必赢的情况就返回赢

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 500;

int n, d;
int dx[maxn], dy[maxn];
int vis[maxn][maxn];

int dfs(int x, int y) {
	if ((x-200)*(x-200) + (y-200)*(y-200) >= d*d)
		return 1;
	if (vis[x][y] != -1)
		return vis[x][y];
	for (int i = 0; i < n; i++)
		if (dfs(x+dx[i], y+dy[i]) == 0)
			return vis[x][y] = 1;
	return vis[x][y] = 0;
}

int main() {
	int x, y;
	scanf("%d%d%d%d", &x, &y, &n, &d);
	x += 200, y += 200;
	for (int i = 0; i < n; i++) 
		scanf("%d%d", &dx[i], &dy[i]);
	memset(vis, -1, sizeof(vis));
	if (dfs(x, y)) 
		printf("Anton\n");
	else printf("Dasha\n");
	return 0;
}