BZOJ 1303: [CQOI2009]中位数图

Posted 2020-10-06 ZlycerQan

二次联通门 : BZOJ 1303: [CQOI2009]中位数图

 

 

 

 

/*
    BZOJ 1303: [CQOI2009]中位数图
    
    对于一个数
    若当前数大于中位数，则记为1
    小于中位数，记为-1，等于则记为0

    对原数列做一个前缀和

    观察发现，若一段区间符合要求
    则该段区间内大于中位数的数的数量等于小于中位数的
    那么统计一下每个前缀和出现的次数就好了
    
*/
#include <cstdio>
#include <iostream>

const int BUF = 12312313;
char Buf[BUF], *buf = Buf;

inline void read (int &now)
{
    for (now = 0; !isdigit (*buf); ++ buf);
    for (; isdigit (*buf); now = now * 10 + *buf - ‘0‘, ++ buf);
}
#define Max 100002
int c[Max], s[Max];
short fc[Max], zc[Max];

int Main ()
{
    fread (buf, 1, BUF, stdin); int N, B; read (N), read (B);
    register int i, j; int x, p = -1, Answer = 0;
    for (i = 1; i <= N; ++ i) 
    {
        read (x); s[i] = s[i - 1];
        if (x > B) ++ s[i];    else if (x < B) -- s[i]; else p = i;
        if (p == -1) s[i] < 0 ? ++ fc[- s[i]] : ++ zc[s[i]];    
    }
    for (++ zc[0]; p <= N; ++ p) Answer += s[p] < 0 ? fc[- s[p]] : zc[s[p]];
    printf ("%d", Answer);    
    return 0;
}
int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}