213 字
1 分钟
中值滤波

题目描述#

20170813082609_40094.jpg

输入输出#

20170813082630_74802.jpg

样例输入#

3
0 1 0

样例输出#

1
0 0 0

提示#

20170813082722_12263.jpg

题解#

先打表

会发现规律然后我们按照规律模拟就可以了

具体的可以看代码

#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int a[500005], b[500005];
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
a[n+1]=a[n];
int i = 1, l, r;
int ans = 0;
while (i <= n)
{
while (i <= n && a[i] == a[i + 1])
{
b[i] = a[i];
i++;
}
l = i;
while (i <= n && a[i] != a[i + 1])
i++;
r = i;
int len = (r - l + 1);
if (len & 1)
for (int j = l; j <= r; j++)
b[j] = a[l];
else
for (int j = 1; j * 2 <= len; j++)
b[l + j - 1] = a[l], b[r - j + 1] = a[r];
ans = max(ans, (len - 1) / 2);
}
printf("%d\n", ans);
for (int i = 1; i <= n; i++)
printf("%d ", b[i]);
//while(1);
}
中值滤波
https://www.nekomio.com/posts/98/
作者
NekoMio
发布于
2017-08-15
许可协议
CC BY-NC-SA 4.0