Codeforces Round #490 (Div. 3) A

传送门http://codeforces.com/contest/999/problem/A

简单地说,就是给你一列数,只能从头和尾开始删小于等于k的数,问最多能删多少个数。

n和k都只有100,暴力题。直接从头扫描一下,删掉能删的;再从后扫描,删掉能删的。记个数,结。

 1 #include <iostream>
 2 using namespace std;
 3 const int maxn = 101;
 4 int n, k, a[maxn], cnt, l;
 5 int main()
 6 {
 7     cin >> n >> k;
 8     for (int i = 1; i <= n; ++i) cin >> a[i];
 9     for (int i = 1; i <= n; ++i)
10     {
11         if (a[i] <= k) ++cnt;
12         else
13         {
14             l = i;
15             break;
16         }
17     }
18     if (cnt == n)
19     {
20         cout << n;
21         return 0;
22     }
23     for (int i = n; i > l; --i)
24     {
25         if (a[i] <= k) ++cnt;
26         else break;
27     }
28     cout << cnt;
29     return 0;
30 }

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转载自www.cnblogs.com/lightgreenlemon/p/9211741.html