``` #include<bits/stdc++.h> using namespace std; int a[] = {0,2,0,2,3}; int ans; int main(){ for(int i = 12345678; i <= 98765432;i++){ int x = i,p = 4; while(x){ int y = x % 10; x /= 10; if(y == a[p]) p--; if(p == 0){ ans++; break; } } } cout << 98765432 - 12345678 - ans + 1; return 0; } ```

Python暴力 ``` import re def count_num(start, end): pattern = re.compile(r'.*2.*0.*2.*3.*') # 正则表达式模式，匹配包含子串"2023"的任意字符 count = 0 for num in range(start, end + 1): if not pattern.search(str(num)): # 如果数字不匹配正则表达式模式 count += 1 return count start = 12345678 end = 98765432 result = count_num(start, end) print(f"在 {start} 至 {end} 中，有 {result} 个数中完全不包含 2023。") ```