2018ICPC Nanjing Station Problem A. Adrien and Austin

Meaning of the questions:

　　the position of the stones the n 1-n on, when two people take turns to be taken by a contiguous stretch, take up the k can not be taken as an input, ask who win

Resolution:

　　When k is 2 or more, the upper hand can always divided into two equal parts rock, no matter how case FLAC go, just get the same strategy can be selected at the symmetrical positions

Code:

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <cstdio>
#include <queue>
#include <cmath>
#include <map>
#include <set>

using namespace std;

typedef long long ll;

int main () {
    ll n,k;
    scanf("%lld%lld",&n,&k);
    if(n==0){
        printf("Austin\n");
    }else if(k==1){
        if(n%2==0){
            printf("Austin\n");
        }else{
            printf("Adrien\n");
        }
    }else {
        printf("Adrien\n");
    }
    return 0;
}