POJ Football Game 【NIMK博弈 && Bash 博弈】

Football Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 451		Accepted: 178

Description

Alice and Bob both love football very much, and both of them are vanguards. They are both good at football control. One day after a football match, they play an interesting game, in which they shoot footballs forward the goal directly. There are N footballs in front of the goal, and they play this game in turn. For example, if it is Alice's turn, Alice can choice some footballs (the number of footballs mush equal or less than M) and shoot them forward. Because the footballs' quality is not very good, footballs are not a complete sphere and can only roll integer times of its girth. And because of restriction of the friction and strength of them, they cannot shoot a football farther then L centimeters. Of course, they know the radius of a football is R centimeters. Alice and Bob love this game very much. If both of them have unlimited IQ and precision shooting skill, can you guess who can win the football game? By the way, though Alice is as strong as Bob, Alice is a girl, so she will shoot first.

Input

The input consists of several cases, each of which contains two lines. 

For each test case, the first line contains 4 integers N, M, L and R (1 <= M <= N <= 30, 0 < L < 100000000, 0 < R < 10000), separated by a single space. N is the number of the footballs, M is the maximum number of footballs one player can shot in one turn, L is the maximum distance that a player can shoot, and R is the radius of footballs. 

The next line contains N numbers, S(1), S(2), ..., S(N) (0 < S(i) < 100000000), which describe the distance between footballs and the goal.

Output

For each case output contains one line describing the name of the winner.

Sample Input

2 1 30 1
8 14
2 1 30 1
8 12
2 1 30 1
8 10
2 1 30 1
40 200

Sample Output

Alice
Bob
Bob
Bob

Source

POJ Monthly,uni

 

 

题意概括：

有 N 个足球每个距离球门 D 远，两人轮流选择小于等于 M 个球射门， 射球最远距离为 L，球的半径为 R （球每次只能移动 周长的整数倍），最后射球者胜，先手赢还是后手赢；

解题思路：

首先要求问题进行转换 N 个球相当于 N 堆物品， 由于每个球只能滚动 周长的整数倍，所以可以把 球到球门的距离 离散化（即选多少个物品），而由于射门有最大距离，所以选取范围 小于 K，（单独看单堆即 Bash 博弈)

接下来考虑 NIM 游戏的变形，普通的 NIM 游戏 每次选取一个堆取若干物品，NIMK 游戏 每次选取 ≤ K 个堆取若干物品。

结论 同样是异或所有堆数，不过不是 模二异或 而是 模(K+1)异或。奇异局势为必败态。 

 

AC code：

//#include <bits/stdc++.h>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define inc(i, j, k) for(int i = j; i <= k; i++)
#define rep(i, j, k) for(int i = j; i < k; i++)
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
#define INF 0x3f3f3f3f
#define LL long long
#define MEM(i, j) memset(i, j, sizeof(i));
#define gcd(i, j) __gcd(i, j)
using namespace std;
const int MAXN = 1e5+10;
const double PI = acos(-1.0);
int SG[MAXN];
int XOR[MAXN];
int xxx;
int num;
int maxn;

bool solve(int N, int M)
{
    MEM(XOR, 0);
    maxn = -1;
    inc(i, 1, N){
        xxx = SG[i];
        num = 0;
        while(xxx){
            XOR[num]+=xxx&1;
            num++;
            xxx>>=1;
        }
        maxn = max(maxn, num);
    }
    rep(i, 0, maxn){
        if(XOR[i]%(M+1)) return true;
    }
    return false;
}
int N, M, L, R;
int s;

int main()
{
    while(~scanf("%d %d %d %d", &N, &M, &L, &R)){
        s = L/(2*PI*R);         //最远的距离
        inc(i, 1, N){
            scanf("%d", &SG[i]);
            SG[i] = SG[i]/(2*PI*R) + 1;   //距离球门的距离
            SG[i]%=s+1;                   // Bash博弈
        }
        if(solve(N, M)) puts("Alice");      //NIMK 博弈
        else puts("Bob");
    }
    return 0;
}