E. Furlo and Rublo and Game
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Furlo and Rublo play a game. The table has n piles of coins lying on it, the i-th pile has ai coins. Furlo and Rublo move in turns, Furlo moves first. In one move you are allowed to:
- choose some pile, let's denote the current number of coins in it as x;
- choose some integer y (0 ≤ y < x; x1 / 4 ≤ y ≤ x1 / 2) and decrease the number of coins in this pile to y. In other words, after the described move the pile will have y coins left.
The player who can't make a move, loses.
Your task is to find out, who wins in the given game if both Furlo and Rublo play optimally well.
Input
The first line contains integer n (1 ≤ n ≤ 77777) — the number of piles. The next line contains n integers a1, a2, ..., an(1 ≤ ai ≤ 777777777777) — the sizes of piles. The numbers are separated by single spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.
Output
If both players play optimally well and Furlo wins, print "Furlo", otherwise print "Rublo". Print the answers without the quotes.
Examples
input
Copy
1 1
output
Copy
Rublo
input
Copy
2 1 2
output
Copy
Rublo
input
Copy
10 1 2 3 4 5 6 7 8 9 10
output
Copy
Furlo
分析：首先枚举出前几项的SG函数值，然后通过尺取法算出后面的SG函数值。
[1,3]的SG为0，SG[4]的函数值为1，考虑左端点l=x^(1/4)和右端点r=x^(1/2)，当且仅当r<4时sg才为1，所以算出sg为1的区间表示为[4,15]，以此类推。由于区间都是平方关系，所以直接手算就可以了。
#include "bits/stdc++.h"
using namespace std;
int sg(long long x)
{
if(x<=3)return 0;
if(x<=15)return 1;
if(x<=81)return 2;
if(x<=6723)return 0;
if(x<=50625)return 3;
if(x<=2562991875LL)return 1;
else return 2;
}
int main() {
int n;
cin >> n;
int ans=0;
for (int i = 1; i <= n; ++i) {
long long x;cin>>x;
ans^=sg(x);
}
if(ans)puts("Furlo");
else puts("Rublo");
}