4351: GESP C++ 四级 4图像压缩202306
内存限制:128 MB
时间限制:1.000 S
评测方式:文本比较
命题人:
提交:1
解决:1
题目描述
样例输入 复制
10
00FFCFAB00FFAC09071B5CCFAB76
00AFCBAB11FFAB09981D34CFAF56
01BFCEAB00FFAC0907F25FCFBA65
10FBCBAB11FFAB09981DF4CFCA67
00FFCBFB00FFAC0907A25CCFFC76
00FFCBAB1CFFCB09FC1AC4CFCF67
01FCCBAB00FFAC0F071A54CFBA65
10EFCBAB11FFAB09981B34CFCF67
01FFCBAB00FFAC0F071054CFAC76
1000CBAB11FFAB0A981B84CFCF66
样例输出 复制
ABCFFF00CB09AC07101198011B6776FC
321032657CD10E
36409205ACC16D
B41032657FD16D
8F409205ACF14D
324F326570D1FE
3240C245FC411D
BF4032687CD16D
8F409205ACC11D
B240326878D16E
83409205ACE11D
提示
#include <iostream>
#include <cstring>
using namespace std;
int image[20][20];
int cpimg[20][20];
int his[256];
int color[16];
// 一位十六进制字符转换为数字
int trans(char a) {
if (a <= '9')
return (a - '0');
return (a - 'A' + 10);
}
// 一位十六进制数字转换为字符
char itrans(int n) {
if (n >= 10)
return (char)(n - 10 + 'A');
return (char)(n + '0');
}
// 寻找离 c 最近的灰阶
int compress(int c) {
int dis = 256, res = -1;
for (int i = 0; i < 16; i++) {
int d = c - color[i];
if (d < 0)
d = -d;
if (d < dis) {
dis = d;
res = i;
}
}
return res;
}
int main() {
int N = 0, M = 0;
cin >> N;
// 灰阶计数,初始化为-1
for (int i = 0; i < 256; i++)
his[i] = -1;
// 输入图像,并对灰阶计数
for (int i = 0; i < N; i++) {
char line[50];
cin >> line;
M = strlen(line) / 2;
for (int j = 0; j < M; j++) {
int c = trans(line[j * 2]) * 16 + trans(line[j * 2 + 1]);
image[i][j] = c;
his[c]++;
}
}
// 选取出现次数最多的 16 个灰阶
for (int c = 0; c < 16; c++) {
int max = 0, max_id = -1;
for (int i = 0; i < 256; i++)
if (his[i] > max) {
max = his[i];
max_id = i;
}
color[c] = max_id;
his[max_id] = -1;
}
// 将 image 的灰阶压缩为 cpimg
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++)
cpimg[i][j] = compress(image[i][j]);
// 输出选取的 16 个灰阶
for (int c = 0; c < 16; c++)
cout << itrans(color[c] / 16) << itrans(color[c] % 16);
cout << endl;
// 输出压缩后的图像
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
cout << itrans(cpimg[i][j]);
cout << endl;
}
return 0;
}