There are n stone quarries in Petrograd.
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 has xi + 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 stones in it.
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».
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 «tolik» if the oligarch who takes a stone first wins, and «bolik» otherwise.
2
2 1
3 2
tolik
4
1 1
1 1
1 1
1 1
bolik
这题是变形的尼姆博弈，知道什么是尼姆博弈的话，难点就在于位运算了。看到一篇博客的代码写的很详细。
附ac代码（有注释）：
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 typedef long long LL; 6 LL get(LL x,LL m) //n^(n+1)=1(n为偶数) 1^1=0 0^0=0 7 { 8 LL ans; 9 if(m&1){ 10 if(x&1) 11 ans=x;//后面的可以配成对 12 else 13 ans=x+m-1;//除去最后一项前面的可以配成对 14 m--; 15 } 16 else{ 17 if(x&1){ 18 ans=x^(x+m-1);//中间的可以配成对 19 m-=2; 20 } 21 else 22 ans=0; 23 } 24 if(m%4)//判断是否为偶数对 25 return ans^1; 26 else 27 return ans; 28 } 29 int main() 30 { 31 int n; 32 long long a,b; 33 while(~scanf("%d",&n)){ 34 long long ans=0; 35 for(int i=0;i<n;i++){ 36 scanf("%lld%lld",&a,&b); 37 ans^=get(a,b); 38 } 39 if(ans) 40 puts("tolik"); 41 else 42 puts("bolik"); 43 } 44 return 0; 45 }
博客地址：http://blog.csdn.net/bigbigship/article/details/31031815