LibreOJ「LibreOJ β Round #4」 游戏

/*
    LibreOJ「LibreOJ β Round #4」 游戏
    
    找找规律就会发现。。
    当有X的时候，答案跟X个数的奇偶有关
    否则就求一下逆序对就好了。。

    由于SB的错误。。WA了3发才过
    然后就签完到走人了
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>

#define Max 11

int key[Max * 10000], rank[Max * 10000];
int s[Max * 10000];

int Main ()
{
    int N, x; scanf ("%d", &N); register int i, j;
    static char l[Max];     int C = 0, cur = 0 ; bool f;
    if (N == 1) return printf ("L"), 0;
    for (i = 1; i <= N; ++ i)
    {
        scanf ("%s", l); f = false, x = 0;
        if (l[0] == ‘X‘) ++ C;
        else
        {
            int Len = strlen (l);
            if (l[0] == ‘-‘) 
            {
                f = true;
                for (j = 1; j < Len; ++ j)
                    if (isdigit (l[j]))
                        x = x * 10 + l[j] - ‘0‘;
            }
            else 
            {
                for (j = 0; j < Len; ++ j)
                    if (isdigit (l[j]))
                        x = x * 10 + l[j] - ‘0‘;
            }
            key[++ cur] = f ? -x : x;
            rank[cur] = key[cur];
        }
    }    
    if (C) return printf (C % 2 ? "W" : "L"), 0;
    long long Answer = 0; int Size;
    std :: sort (key + 1, key + 1 + cur);
    Size = std :: unique (key + 1, key + 1 + cur) - key - 1;
    for (i = 1; i <= cur; ++ i)
    {
        rank[i] = std :: lower_bound (key + 1, key + 1 + Size, rank[i]) - key;
        int res = 0;
        for (j = rank[i]; j <= cur; j += j & -j) ++ s[j];
        for (j = rank[i]; j; j -= j & -j) res -= s[j];
        for (j = cur; j; j -= j & -j) res += s[j];
        Answer += res;
    }
    printf (Answer % 2 ? "W" : "L");

    return 0;
}

int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}