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; }