20171108校内模拟赛

 

 

哭死，lqz已经没救了。

期望得分: 100 + 0 + 100

实际得分: 50   + 0 +  0 / 30

 

 

 

T1 有个本应该是int的数组开成了char，挂了50分。

 

T2 没做

 

T3一共写了三个不同的做法，最稳的一个一直到收代码后5min才调出来，考试的时候随便交了一个上去，结果linux下CE，Windows下30

 

/*
    模拟每次得分情况就好
        madan坑我50分去
*/
#include <cstdio>
#include <iostream>

#define Max 100
char a[Max]; int f[Max];
#define rg register
inline void cmax (int &a, int b) { if (b > a) a = b; }
std :: string Name = "soccer", _I = ".in", _O = ".out";
int main (int argc, char *argv[])
{
    freopen ((Name + _I).c_str (), "r", stdin);
    freopen ((Name + _O).c_str (), "w", stdout);
    int N; scanf ("%d", &N); rg int i, j;
    for (i = 1; i <= N; ++ i)
    {
        scanf ("%s", a + 1);
        for (j = 1; j <= N; ++ j)
            if (a[j] == 'W') f[i] += 3;
            else if (a[j] == 'L') f[j] += 3;
            else if (a[j] == 'D') ++ f[i], ++ f[j];
    }
    int Maxn = 0;
    
    for (i = 1; i <= N; ++ i) cmax (Maxn, f[i]);
    for (i = 1; i <= N; ++ i)
        if (f[i] == Maxn) printf (i == N ? "%d" : "%d ", i); 
    
    return 0;
}

 

 

并不想改T2

 

/*
    set 维护一下就好
    考场上基本想到了标算
    但是维护最长空位置的数据结构我选的是堆
    比较难调
    set就好多了
    
    每次取出一段，取中点，然后把这段拆成两半，插入回set取
     
    把车开走就需要再维护一个set，存每次拆出的点。 
*/
#include <cstdio>
#include <iostream>
#include <set>
int N;
#define rg register
inline void read (int &n)
{
    rg char c = getchar ();
    for (n = 0; !isdigit (c); c = getchar ());
    for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
}
#define Max 1000006
struct D
{
    int l, r; D (int a = 0, int b = 0) : l (a), r (b) {} 
    inline int gl () const
    {
        if (l == 1) return r;
        return r == N ? r - l + 1 : (l + r) / 2 - l + 1;
    }
    
    bool operator < (const D &rhs)  const 
    { return gl () == rhs.gl () ? l < rhs.l : (gl () > rhs.gl ()); }
    
};
std :: set <D> A; std :: set <int> B;
int a[Max];
int main (int argc, char *argv[])
{
    freopen ("park.in", "r", stdin);
    freopen ("park.out", "w", stdout);
    int M; read (N), read (M); rg int i;
#define ir insert
    A.ir (D (1, N)), B.ir (N + 1), B.ir (0);
    D n, p; int t, x, e;
    std :: set <int> :: iterator it;
    for (; M; -- M)
    {
        read (t), read (x);
        if (t == 1)
        {
            n = *A.begin (), A.erase (n);
            if (n.l == 1) e = 1, n.l = 2, A.ir (n);
            else if (n.r == N) e = N, n.r = N - 1, A.ir (n);
            else
                e = n.l + n.r >> 1, p = n, n.r = e - 1, p.l = e + 1, A.ir (p), A.ir (n);
            printf ("%d\n", a[x] = e); B.ir (e); continue;
        }
        it = B.find (a[x]);
        n.l = *(-- it) + 1, n.r = a[x] - 1, A.erase (n);
        p.l = a[x] + 1, p.r = *(++++ it) - 1, A.erase (p);
        n.r = p.r, A.ir (n); B.erase (a[x]), a[x] = 0;
    }
    return 0;
}