链接：https://www.nowcoder.com/acm/contest/125/A
来源：牛客网
Tony and Macle are good friends. One day they joined a birthday party together. Fortunately, they got the opportunity to have dinner with the princess, but only one person had the chance. So they decided to play a game to decide who could have dinner with the princess.
The rules of the game are very simple. At first, there are n sticks. Each of exactly m meters in length. Each person takes turns. For one move the player chooses just one stick and cut it into several equal length sticks, and the length of each stick is an integer and no less than k. Each resulting part is also a stick which can be chosen in the future. The player who can’t make a move loses the game ,thus the other one win. Macle make the first move.
输入描述:
the first line contains tree integer n,m,k(1 <= n,m,k <= 10
9
)
输出描述:
print 'Macle' if Macle wins or 'Tony' if Tony wins,you should print everything without quotes.
<!–
–>
输入例子:
1 15 4
输出例子:
Macle
–>
示例1
输入
1 15 4
输出
Macle
示例2
输入
4 9 5
输出
Tony有n根m长的绳子，每次你可以把任意一根绳子分成不小于k长度的任意根绳子。分到的绳子在下次可以被分解。Macle先开始，谁先不能对这些绳子进行操作就输了。
如果n是偶数的话，那么Macle肯定输了。Macle要赢的条件是：(1）n是奇数的(2)m有小于k的因子，这样我就可以一次性把m分解成一个Tony无法再分解的数
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<cmath>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll inf = 1e9;
const ll maxn = 1e3 + ;
ll gcd( ll a, ll b ) {
if( a == ) {
return b;
}
if( b == ) {
return a;
}
return gcd( b, a%b );
}
int main() {
std::ios::sync_with_stdio(false);
ll n, m, k;
while( cin >> n >> m >> k ) {
ll num = m / k;
bool flag = false;
for( ll i = num; i > ; i -- ) {
if( m % i == ) {
flag = true;
break;
}
}
if( ( n & ) && flag ) {
cout << "Macle" << endl;
} else {
cout << "Tony" << endl;
}
}
return ;
}
