參考：https://www.cnblogs.com/qscqesze/p/6628793.html
A. Berzerk
題目連接：
http://codeforces.com/contest/786/problem/A
Description
Rick and Morty are playing their own version of Berzerk (which has nothing in common with the famous Berzerk game). This game needs a huge space, so they play it with a computer.
In this game there are n objects numbered from 1 to n arranged in a circle (in clockwise order). Object number 1 is a black hole and the others are planets. There’s a monster in one of the planet. Rick and Morty don’t know on which one yet, only that he’s not initially in the black hole, but Unity will inform them before the game starts. But for now, they want to be prepared for every possible scenario.
Each one of them has a set of numbers between 1 and n - 1 (inclusive). Rick’s set is s1 with k1 elements and Morty’s is s2 with k2 elements. One of them goes first and the player changes alternatively. In each player’s turn, he should choose an arbitrary number like x from his set and the monster will move to his x-th next object from its current position (clockwise). If after his move the monster gets to the black hole he wins.
Your task is that for each of monster’s initial positions and who plays first determine if the starter wins, loses, or the game will stuck in an infinite loop. In case when player can lose or make game infinity, it more profitable to choose infinity game.
Input
The first line of input contains a single integer n (2 ≤ n ≤ 7000) — number of objects in game.
The second line contains integer k1 followed by k1 distinct integers s1, 1, s1, 2, …, s1, k1 — Rick’s set.
The third line contains integer k2 followed by k2 distinct integers s2, 1, s2, 2, …, s2, k2 — Morty’s set
1 ≤ ki ≤ n - 1 and 1 ≤ si, 1, si, 2, …, si, ki ≤ n - 1 for 1 ≤ i ≤ 2.
Output
In the first line print n - 1 words separated by spaces where i-th word is “Win” (without quotations) if in the scenario that Rick plays first and monster is initially in object number i + 1 he wins, “Lose” if he loses and “Loop” if the game will never end.
Similarly, in the second line print n - 1 words separated by spaces where i-th word is “Win” (without quotations) if in the scenario that Morty plays first and monster is initially in object number i + 1 he wins, “Lose” if he loses and “Loop” if the game will never end.
Sample Input
5
2 3 2
3 1 2 3
Sample Output
Lose Win Win Loop
Loop Win Win Win
Hint
題意
有兩個人在一個環上玩遊戲，由n個格子組成的環，其中0環是洞。
現在A，B兩個人各自擁有K[i]個選項，第i個選項是讓怪獸順時針走s[i]步。
兩個人輪流讓怪獸走，誰讓怪獸走進洞裏面誰就勝利。
現在問你考慮所有情況，勝利的結果是什麼。
題解：
記憶化搜索。
倒着來。dp[i][j]表示現在i先手位置在j的勝負情況。
顯然如果轉移到dp[i][j]的狀態全是對手勝利的話，那麼dp[i][j]就是失敗。
如果轉移到dp[i][j]的狀態存在對手失敗，那麼dp[i][j]就是勝利。
其他都是無限循環。
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define N 8010
string ans[3]={"Lose","Loop","Win"};
int n;
vector<int>Move[2];
int Flag[2][N];
int cnt[2][N];
void dfs(int x,int y,int val){
Flag[x][y]=val;
if(val==-1){
for(int i=0;i<Move[!x].size();++i){
if(!Flag[!x][(y+n-Move[!x][i])%n])
dfs(!x,(y+n-Move[!x][i])%n,1);
}
}
else{
for(int i=0;i<Move[!x].size();++i){
if(Flag[!x][(y+n-Move[!x][i])%n])continue;
if(cnt[!x][(y+n-Move[!x][i])%n]<Move[!x].size()){
cnt[!x][(y+n-Move[!x][i])%n]++;
}
if(cnt[!x][(y+n-Move[!x][i])%n]==Move[!x].size()){
dfs(!x,(y+n-Move[!x][i])%n,-1);
}
}
}
}
int main(){
//freopen("in.txt","r",stdin);
scanf("%d",&n);
for(int tp=0;tp<2;++tp){
int k;
scanf("%d",&k);
for(int i=0;i<k;++i){
int tmp;
scanf("%d",&tmp);
Move[tp].push_back(tmp);
}
}
Flag[1][0]=-1;
dfs(0,0,-1);
dfs(1,0,-1);
for(int tp=0;tp<2;++tp){
for(int i=1;i<n;++i){
cout<<ans[Flag[tp][i]+1]<<' ';
}puts("");
}
}