A
After Asgard was destroyed, tanker brought his soldiers to earth, and at the same time took on the important task of protecting the peace of the earth. The best two solders were lb and zgx, were very capable, but they always disliked each other. However, one day they encountered a group of foreign invaders (many, but how many only tanker knew). They were all strong enough to destroy the enemy easily. But they found it too boring, so they agreed to follow some rules to deal with the invaders by taking turns, and if one of them had no enemies when it was his turn, he would later admit that the other man was better.
The rules are as follows:
To ensure fairness, they found their leader, tanker, to judge, but tanker just wanted people to say he was great, so he didn't want them to decide easily, so he hid the number of intruders in a question:
In the first line, input kk, and on lines 22 to k + 1k+1, input kk groups aa and bb.
If lb wins, output "Lbnb!
", if zgx wins, output "Zgxnb!
", if they can't solve, (nn does not exist) , output "Tankernb!
" .
k\le 10k≤10 ,1< n \le 10^{15}n≤1015
For the sample, n=8n=8,because 8\%5=38%5=3, 8 \%3=28%3=2 and 88 is the smallest possible integer that is fit the requirement.
样例输入复制
2
5 3
3 2
样例输出复制
Lbnb!
扩展式中国剩余定理+斐波那契博弈,先打表找规律,找到必输的情况。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
#define LL long long
LL mi[1100],ai[1100],fb[1100];//mi为要模的数,ai为余数。
map<LL,bool>p;
LL gcd(LL a, LL b)
{
return b == 0 ? a : gcd(b, a%b);
}
void exgcd(LL a, LL b, LL &d, LL &x, LL &y)
{
if(!b)
{
d = a, x = 1, y = 0;
}
else
{
exgcd(b, a%b, d, y, x);
y -= x * (a / b);
}
}
LL CRT(LL l, LL r, LL *mi, LL *ai)
{
LL lcm = 1;
for(LL i = l; i <= r; i++)
lcm = lcm / gcd(lcm, mi[i]) * mi[i];
for(LL i = l+1; i <= r; i++)
{
LL A = mi[l], B = mi[i], d, x, y, c = ai[i] - ai[l];
exgcd(A, B, d, x, y);
if(c % d)
return -1;
LL mod = mi[i] / d;
LL k = ((x * c / d) % mod + mod) % mod;
ai[l] = mi[l] * k + ai[l];
mi[l] = mi[l] * mi[i] / d;
}
if(ai[l] == 0)
return lcm;
return ai[l];
}
int main()
{
LL t,n,i;
scanf("%lld",&t);
for(i=1; i<=t; i++)
scanf("%lld%lld",&mi[i],&ai[i]);
n=CRT(1ll,t,mi,ai);
if(n>1e15||n==-1)
{
printf("Tankernb!");
return 0;
}
fb[1]=2,fb[2]=3;
p[2]=true,p[3]=true;
for(i=3;i<=320;i++)
{
fb[i]=fb[i-1]+fb[i-2];
if(fb[i]>1e15)
break;
p[fb[i]]=true;
}
if(p[n])
printf("Lbnb!");
else printf("Zgxnb!");
return 0;
}