Educational Codeforces Round 68 (Rated for Div. 2)D（SG函数打表，找规律）

#include<bits/stdc++.h>
using namespace std;
int sg[1007];
int main(){
int t;
cin>>t;
while(t--){
int n,k;
cin>>n>>k;
if(k%3==0){
n%=(k+1);
if(n==k||n%3)
cout<<"Alice"<<"\n";
else
cout<<"Bob"<<"\n";
}
else{
if(n%3)
cout<<"Alice"<<"\n";
else
cout<<"Bob"<<"\n";
}
/*
cin>>n>>k;
sg[1]=1;
sg[2]=1;
sg[k]=1;
for(int i=3;i<=n;++i){
if(i>=k){
if(sg[i-1]==0||sg[i-2]==0||sg[i-k]==0)
sg[i]=1;
}
else{
if(sg[i-1]==0||sg[i-2]==0)
sg[i]=1;
}
}
for(int i=0;i<=n;++i){
cout<<sg[i]<<"\n";
}*/
//sg函数打表找规律，发现当k是3的倍数时，sg函数以k+1为循环节，当k不是3的倍数时，sg函数以3为循环节
}
return 0;
}