Codeforces 786A Berzerk DP+博弈

A. Berzerk

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

52 3 23 1 2 3

output

Lose Win Win LoopLoop Win Win Win

input

84 6 2 3 42 3 6

output

Win Win Win Win Win Win WinLose Win Lose Lose Win Lose Lose

一个环上N个位置。两个人轮流玩游戏，每次可以从各自的集合中选择任意一个数作为自己的步长，走到1号点就输了。问把2到n号位置作为出发点，两人中各自先手共2*(N-1)种情况中，最终先手会获胜、失败还是平局。

这题，直接从每一种情况向前推肯定行不通，因为情况太多了。

这样来看，从必败态1号点逆推似乎比较简单。

接着按照博弈论中的理论从1号点向前逆推。可达到必败态的点，一定是必胜点，而所有后续状态全为对方必胜态的点则为必败点。每次从集合中选数，将新状态加入队列即可。

用1和2表示上一步是谁走的。

怎么判断是否全部的后续状态都是对方必胜态？可以开两个数组，大小与圆环长度相同，最开始的值为自己集合内数的个数，每推到一个就将值-1，值为0代表所有状态都被推完。

#include <cstdio>#include <iostream>#include <string.h>#include <queue>using namespace std;const int maxn=7005;int dp[maxn],dp2[maxn],a[maxn],b[maxn],r[maxn],r2[maxn];struct node {int pos,pre;node(int pos,int pre): pos(pos),pre(pre) {}};int main() {int n,k1,k2,i,h=0,t=0,cnt,j;scanf("%d%d",&n,&k1);memset(dp,0,sizeof(dp));memset(dp2,0,sizeof(dp2));for (i=1;i<=k1;i++) {scanf("%d",&a[i]);}dp[0]=2;scanf("%d",&k2);int h2=0,t2=0;for (i=1;i<=k2;i++) {scanf("%d",&b[i]);}for (i=1;i<=n;i++) {r[i]=k1;r2[i]=k2;}dp2[0]=2;cnt=1;queue<node> q;q.push(node(0,1));q.push(node(0,2));while (!q.empty()) {int now=q.front().pos,p=q.front().pre;q.pop();if (p==1) {for (i=1;i<=k2;i++) {int pos=now-b[i];if (pos<0) pos+=n;if (!dp2[pos]) {if (dp[now]==2) {dp2[pos]=1;q.push(node(pos,2));} else {r2[pos]--;if (r2[pos]==0) {dp2[pos]=2;q.push(node(pos,2));}}}}} else {for (i=1;i<=k1;i++) {int pos=now-a[i];if (pos<0) pos+=n;if (!dp[pos]) {if (dp2[now]==2) {dp[pos]=1;q.push(node(pos,1));} else {r[pos]--;if (r[pos]==0) {dp[pos]=2;q.push(node(pos,1));}}}}}}for (i=1;i<=n-1;i++) {if (dp[i]==0) printf("Loop "); else    if (dp[i]==1) printf("Win "); else printf("Lose ");}printf("\n");for (i=1;i<=n-1;i++) {if (dp2[i]==0) printf("Loop "); else    if (dp2[i]==1) printf("Win "); else printf("Lose ");}return 0;}