Description
There are n
Each quarry owns mi dumpers (1 ≤ i ≤ n). It is known that the first dumper of the i-th quarry has xi stones in it, the second dumper hasxi + 1 stones in it, the third has xi + 2, and the mi-th dumper (the last for the i-th quarry) has xi + mi - 1
Two oligarchs play a well-known game Nim. Players take turns removing stones from dumpers. On each turn, a player can select any dumper and remove any non-zero amount of stones from it. The player who cannot take a stone loses.
Your task is to find out which oligarch will win, provided that both of them play optimally. The oligarchs asked you not to reveal their names. So, let's call the one who takes the first stone «tolik» and the other one «bolik».
Input
The first line of the input contains one integer number n (1 ≤ n ≤ 105) — the amount of quarries. Then there follow n lines, each of them contains two space-separated integers xi and mi (1 ≤ xi, mi ≤ 1016) — the amount of stones in the first dumper of the i-th quarry and the number of dumpers at the i-th quarry.
Output
Output «tolik» if the oligarch who takes a stone first wins, and «bolik» otherwise.
Sample Input
Input
2
2 1
3 2
Output
tolik
Input
4
1 1
1 1
1 1
1 1
Output
bolik
nim取子游戏，异或和为0即为先手必败，其余为先手必胜，然后问题转换为求l到r的异或和，再转化为求1到n的异或和。打表找找规律可以知道只有四种情况。
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<string>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
#define LL __int64
int T;
LL get(LL x)
{
if ((x&3)==0) return x;
if ((x&3)==1) return 1;
if ((x&3)==2) return x+1;
if ((x&3)==3) return 0;
}
int main()
{
while (~scanf("%d",&T))
{
LL ans=0,x,y;
while(T--)
{
scanf("%I64d%I64d",&x,&y);
ans^=get(x-1)^get(x+y-1);
}
if (ans) printf("tolik\n");
else printf("bolik\n");
}
return 0;
}