LibreOJ #524. 「LibreOJ β Round #4」游戏

二次联通门 : LibreOJ #524. 「LibreOJ β Round #4」游戏

/*
    LibreOJ #524. 「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 * ], rank[Max * ];
int s[Max * ];

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

    return ;
}

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