【codeforces 768A】Oath of the Night's Watch

【题目链接】:http://codeforces.com/contest/768/problem/A

【题意】



让你统计这样的数字x的个数;

x要满足有严格比它小和严格比它大的数字;

【题解】



排个序,把最左边和最右边的元素剔除掉;

中间剩下的就是满足要求的元素了;



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1e5+100;

int a[N];
int n;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
        rei(a[i]);
    sort(a+1,a+1+n);
    int l = 1,r = n;
    while (l+1<=n && a[l+1]==a[1]) l++;
    while (r-1>=1 && a[r-1]==a[n]) r--;
    int ans = 0;
    rep1(i,l+1,r-1)
        ans++;
    printf("%d\n",ans);
    return 0;
}