D. 1-2-K Game
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Alice and Bob play a game. There is a paper strip which is divided into n + 1 cells numbered from left to right starting from 0. There is a chip placed in the n-th cell (the last one).

Players take turns, Alice is first. Each player during his or her turn has to move the chip 1, 2 or k cells to the left (so, if the chip is currently in the cell i, the player can move it into cell i - 1, i - 2 or i - k). The chip should not leave the borders of the paper strip: it is impossible, for example, to move it k cells to the left if the current cell has number i < k. The player who can't make a move loses the game.

Who wins if both participants play optimally?

Alice and Bob would like to play several games, so you should determine the winner in each game.

Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of games. Next T lines contain one game per line. All games are independent.

Each of the next T lines contains two integers n and k (0 ≤ n ≤ 109, 3 ≤ k ≤ 109) — the length of the strip and the constant denoting the third move, respectively.

Output
For each game, print Alice if Alice wins this game and Bob otherwise.

Example
inputCopy
4
0 3
3 3
3 4
4 4
outputCopy
Bob
Alice
Bob
Alice

Title:
At present, in the n position, you can take 1, 2, or k steps to the left every time. The leftmost position is 0. When you can't go to 0, the two players play a game. Who hasn't sent out and who loses, ask whether the first player will win or the second player will win.
Train of thought:

First of all, it is determined that the position of 0 is a required number position, because the three positions of 12 and k can go to the position of 0 in one step, so these three positions are the required winning positions. According to this law, we can recursively deduce the sg function.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int sg[maxn];
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\code_stream\\out.txt","w",stdout);
    
    int n,k;
    n=100;
    cin>>k;
    sg[0]=0;
    sg[1]=1;
    sg[2]=1;
    repd(i,3,n)
    {
        if((i-k)>=0)
        {
            if(sg[i-1]==0||sg[i-2]==0||sg[i-k]==0)
            {
                sg[i]=1;
            }
        }else
        {
            if(sg[i-2]==0||sg[i-1]==0)
            {
                sg[i]=1;
            }
        }
    }
    repd(i,0,n)
    {
        cout<<i<<" "<<sg[i]<<endl;

    }
    
    
    
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

Through the analysis of sg function, we can get the following rules
 If K is a multiple of 3, then sg function is a cyclic section of k+1 length. After moduling the cyclic section, judge whether n is k, if K, then K position must win, otherwise judge whether it is a multiple of 3.
If k is not a multiple of 3, then judge whether n is a multiple of 3.

I read the wrong table and put forward a wrong rule in the game. I have wa three times. I want to write this blog late at night to give myself a memory.

See code for details:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
 
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\code_stream\\out.txt","w",stdout);
 
    int t;
    ll n, k;
    gbtb;
    cin >> t;
    while (t--)
    {
        cin >> n >> k;
        if (k % 3 == 0)
        {
            n %= (k + 1);
            if (n == k)
            {
                cout << "Alice" << endl;
            } else
            {
                if (n % 3 == 0)
                    cout << "Bob" << endl;
                else
                    cout << "Alice" << endl;
            }
            // return 0;
            continue;
        }
 
        if ((n % 3) == 0 )
        {
            cout << "Bob" << endl;
        } else
        {
            cout << "Alice" << endl;
        }
    }
 
 
 
    return 0;
}
 
inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}