codeforces - 15C Industrial Nim(位运算+尼姆博弈)

最新推荐文章于 2020-09-30 10:22:10 发布

转载最新推荐文章于 2020-09-30 10:22:10 发布 · 194 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zmin/p/8490701.html

本文解析了一道关于工业背景下的尼姆博弈问题。玩家轮流从不同容量的石堆中取石子，先无法取者输。文章通过位运算技巧提供了一个高效的解决方案，并附带AC代码及详细解释。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

C. Industrial Nim

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

There are n stone quarries in Petrograd.

Each quarry owns m_i dumpers (1 ≤ i ≤ n). It is known that the first dumper of the i-th quarry has x_i stones in it, the second dumper has x_i + 1 stones in it, the third has x_i + 2, and the m_i-th dumper (the last for the i-th quarry) has x_i + m_i - 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».

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 x_i and m_i (1 ≤ x_i, m_i ≤ 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.

Examples

Input

Copy

2
2 1
3 2

Output

tolik

Input

Copy

Output

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 }