Codeforces Gym - 101498I Rock Piles 博弈

本文介绍了一款名为RockPiles的游戏,该游戏由两名玩家轮流从两个石头堆中取走石头,目标是让对手无法继续操作。文章给出了一个简单的算法来判断在最优策略下哪位玩家将获胜。

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I. Rock Piles
time limit per test
1.0 s
memory limit per test
256 MB
input
standard input
output
standard output

Hasan and Abdullah are lost in the desert, they are waiting for the rescue squad to help them. While they are waiting, they decided to invent a new game called Rock Piles.

In this game, there are two rock piles. The first pile contains N rocks, and the second one contains M rocks. Hasan and Abdullah take turns to play the game, Hasan starts first. In each turn, a player can choose to take one rock from a single pile, or take one rock from both piles. The player who cannot make any move, loses the game. Determine who will win if both of them play optimally.

Input

The first line of the input contains an integer T (1  ≤  T  ≤  104), where T is the number of the test cases.

Each test case has one line that contains two integers N and M (0  ≤  NM  ≤  109), the number of rocks in the first and second pile respectively.

Output

For each test case, print a single line with the winner's name. If Hasan wins print "hasan", otherwise print "abdullah" (without quotes).

Example
input
2
2 3
4 0
output
hasan
abdullah

 

 威佐夫博弈的变形  变形了那跟原来的关系应该就不大了 在纸上写了几个状态没发现规律 只好SG函数打表

打出表后规律一下就清楚了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int, int> PII;
using namespace std;

const int maxn = 10000 + 5;
/*
void GetSG(int n = 100, int m = 100) {
    int i, j;
    memset(SG, 0, sizeof(SG));
    for(i = 0; i <= n; i++) {
        for(int j = 0; j <= m; j++) {
            memset(Hash, 0, sizeof(Hash));
            if(i >= 1) Hash[SG[i - 1][j]] = 1;
            if(j >= 1) Hash[SG[i][j - 1]] = 1;
            if(i >= 1 && j >= 1) Hash[SG[i - 1][j - 1]] = 1;
            for(int k = 0; k <= n; k++) {
                if(Hash[k] == 0) {
                     SG[i][j] = k;
                     break;
                }
            }
        }
    }
}
*/
int main() {
    //FIN
    int n, m;
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        if(n % 2 == 0 && m % 2 == 0) puts("abdullah");
        else puts("hasan");
    }

    return 0;
}

  

转载于:https://www.cnblogs.com/Hyouka/p/7445986.html

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