D. Game with Powers

Vasya and Petya wrote down all integers from 1 to n to play the "powers" game (n can be quite large; however, Vasya and Petya are not confused by this fact).

Players choose numbers in turn (Vasya chooses first). If some number x is chosen at the current turn, it is forbidden to choose x or all of its other positive integer powers (that is, x2, x3, ...) at the next turns. For instance, if the number 9 is chosen at the first turn, one cannot choose 9 or 81 later, while it is still allowed to choose 3 or 27. The one who cannot make a move loses.

Who wins if both Vasya and Petya play optimally?

Input

Input contains single integer n (1 ≤ n ≤ 109).

Output

Print the name of the winner — "Vasya" or "Petya" (without quotes).

Examples

input

1

output

Vasya

input

2

output

Petya

input

8

output

Petya

Note

In the first sample Vasya will choose 1 and win immediately.

In the second sample no matter which number Vasya chooses during his first turn, Petya can choose the remaining number and win.

 我感觉这个问题，要是深入的了解了SG函数做这个题是很轻松的，但是我菜啊，so只能先把别人的代码看懂了，但是真的感觉这个问题是SG函数的经典的问题。因为要是把SG函数写在里面真的是时间爆炸啊，因为集合中最多才可能是29个，所以直接打表然后就OK了

#include
using namespace std;
long long cnt;
int vis[200000];
mapf;
/*int SG(long long r)
{
    int vis[40];
    memset(vis,0,sizeof(vis));
    if(f[r]) return f[r];
    if(r==0) return 0;
    for(int i=1;i<=30;i++)
    {
        if(r&(1<<(i-1)))
        {
            long long tem=r;
            for(int j=i;j<=30;j+=i)
            {
                cnt++;
                if(tem&(1<<(j-1)))
                tem^=(1<<(j-1));
            }
            vis[SG(tem)]=1;
        }
    }
    for(int i=0;;i++)
    {
        if(vis[i]==0)
        {
            f[r]=i;
            return i;
        }
    }
}*/
int main()
{
    int sg[30]={0, 1, 2, 1, 4, 3, 2, 1, 5, 6, 2, 1, 8, 7, 5, 9, 8, 7, 3, 4, 7, 4, 2, 1, 10, 9, 3, 6, 11, 12};
    int n;
    scanf("%d",&n);
    int k;
    for(k=1;;k++)
    {
        if(k*k>=n) break;
    }
    f[1]=1;
    long long sum=0;
    long long ans=0;
    for(int i=2;i<=k;i++)
    {
        int cnt=0;
        long long temp=i;
        if(vis[temp]) continue;
        while(temp<=n)
        {
            cnt++;
            if(temp<=k) vis[temp]=1;
            if(temp>k) sum++;
            temp*=i;
        }
        f[cnt]++;
    }
    f[1]+=n-k-sum;
    for(int i=1;i<=30;i++)
    {
        if(f[i]&1) ans^=sg[i];
    }
    if(ans) printf("Vasya");
    else printf("Petya");
}

　　

转载于:https://www.cnblogs.com/Heilce/p/6582131.html