- 图像模糊处理
AC题解
- @ 2026-4-19 13:12:12
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#include<bits/stdc++.h> using namespace std; int n,m,a[102][102],dx[]={-1,1,0,0},dy[]={0,0,-1,1},ans[102][102]; typedef double d; int f(d dup,d ddw,d dlf,d drg,d dij){ int x=round((dup+ddw+dlf+drg+dij)/5.0); return x; } int main(){ cin>>n>>m; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin>>a[i][j]; for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(i==1||j==1||i==n||j==m){ ans[i][j]=a[i][j]; }else{ d dij=(d)(a[i][j]*1.0),dup=(d)(a[i+dx[0]][j+dy[0]]),ddw=(d)(a[i+dx[1]][j+dy[1]]),dlf=(d)(a[i+dx[2]][j+dy[2]]),drg=(d)(a[i+dx[3]][j+dy[3]]); ans[i][j]=f(dup,ddw,dlf,drg,dij); } } } for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ cout<<ans[i][j]<<" "; } cout<<endl; } return 0; } -
#include <bits/stdc++.h>
using namespace std;
int a[110][110], b[110][110], n, m;
int main() {
cin >> n >> m; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { cin >> a[i][j]; b[i][j] = a[i][j]; } } for (int i = 2; i <= n - 1; i++) { for (int j = 2; j <= m - 1; j++) { b[i][j] = round((a[i][j] + a[i - 1][j] + a[i + 1][j] + a[i][j - 1] + a[i][j + 1]) / 5.0); } } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { cout << b[i][j] << " "; } cout << endl; } return 0;}
- 1
信息
- ID
- 1403
- 时间
- ms
- 内存
- MiB
- 难度
- 2
- 标签
- 递交数
- 179
- 已通过
- 85
- 上传者