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

2019-11-08 01:26:00

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【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;}

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