chess

一道博弈题，可以先用SG函数打表，看看什么时候先手会输。
发现(1,2),(3,5),(4,7)...其实就是个威佐夫博弈。
实际上向左边移动等价于取第一堆石子，向下移动等价于取第二堆石子，向左下移动等价于两堆同时取
相同数目的石子。

#include <stdio.h>
#include <utility>
#include <vector>
#include <string.h>
#include <math.h>
#include <iostream>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#include <queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

#ifdef LOCAL
#define debug(x) cout << "[" __FUNCTION__ ": " #x " = " << (x) << "]\n"
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
#define hash_ 1000000009
#define Continue(x) { x; continue; }
#define Break(x) { x; break; }
const int mod = 998244353;
const int N = 2e4 + 10;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
#define gc p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;
inline int read(){ static char buf[1000000], *p1 = buf, *p2 = buf; register int x = false; register char ch = gc; register bool sgn = false; while (ch != '-' && (ch < '0' || ch > '9')) ch = gc; if (ch == '-') sgn = true, ch = gc; while (ch >= '0'&& ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc; return sgn ? -x : x; }
ll fpow(ll a, int b, int mod) { ll res = 1; for (; b > 0; b >>= 1) { if (b & 1) res = res * a % mod; a = a * a % mod; } return res; }
int dir[3][2] = { -1, -1, -1, 0, 0, -1 };
int d[1010][1010];
void dfs(int x, int y)
{
    if (d[x][y] != -1)
        return ;
    int flag = 1;
    for (int i = 0; i < 3; i++)
    {
        int flag = 1;
        int nx = x, ny = y;
        nx += dir[i][0], ny += dir[i][1];
        while (nx >= 0 && nx <= 100 && ny >= 0 && ny <= 100)
        {
            dfs(nx, ny), flag &= d[nx][ny];
            nx += dir[i][0], ny += dir[i][1];
        }
        if (!flag)
        {
           d[x][y] = 1;
           return;
        }
    }
    d[x][y] = 0;
    return;
}
int main()
{
#ifdef LOCAL
    freopen("E:/input.txt", "r", stdin);
#endif
    /*
    memset(d, -1, sizeof d);
    d[1][0] = d[0][1] = d[1][1] = 1;
    d[0][0] = 0;
    for (int i = 0; i <= 100; i++)
        for (int j = 0; j <= 100; j++)
            dfs(i, j);
    for (int i = 0; i <= 20; i++)
        for (int j = 0; j <= 20; j++)
            if (d[i][j] == 0)
                cout << i << " " << j << " " << "Lao Wang" << endl;
    */
    int a, b;
    double r = (sqrt(5) + 1) / 2;
    while (~scanf("%d%d", &a, &b))
    {
        if (a > b)
            swap(a, b);
        int result = (b - a) * r;
        if (result == a)
            puts("Lao Wang");
        else
             puts("Xiao Ren");
    }
    return TIME;
}