华为笔试 华为笔试题 0424
笔试时间：2025年04月24日
历史笔试传送门：
第一题
题目：物料回收计算器
某云站点有一批服务器待下线，已知这些服务器所对应的物料编码(代表不同服务器型号的标识字符串)，特定物料编码本身已包含CPU/内存/主板等备件型号，请计算出这批服务器下线后可回收不同的CPU/内存/主板型号和数量。物料编码只包含大写英文字母和数字，其中一定包含三类备件型号，它的固定格式是一个大写字母+两个数字，字母C代表CPU，M代表内存Memory，B代表主板Board。其他字母和数字可忽略，同类备件存在多个型号时取第一个。
输入描述
第一行为服务器个数N，范围是[1，1000]:
第二行为N个物料编码字符串，依次用空格隔开
输出描述
统一按照型号1,双量1,型号2,数量2,型号3,数量3.的格式输出
第一行为CPU的型号和对应数量
第二行为内存的型号和对应数量
第三行为支板的型号和对应数量
同类备件存在多个型号时按照字典序输出，分号隔开
样例输入
2
C01M23B050130 C01M23B060130
样例输出
C01,2
M23,2
B05,1;B06,1
参考题解
直接模拟，注意字符串拆分，以及对同类备件多次出现的时候特殊处理一下即可。
C++：[此代码未进行大量数据的测试，仅供参考]
#include <bits/stdc++.h>
using namespace std;
// Parse out first C##, M##, B## segments
tuple<string,string,string> parse_code(const string &code) {
string cpu="", memory="", board="";
int L = code.size();
for(int i = 0; i + 2 < L; ++i) {
char t = code[i];
if (t=='C' && isdigit(code[i+1]) && isdigit(code[i+2])) {
if (cpu.empty()) cpu = code.substr(i,3);
}
else if (t=='M' && isdigit(code[i+1]) && isdigit(code[i+2])) {
if (memory.empty()) memory = code.substr(i,3);
}
else if (t=='B' && isdigit(code[i+1]) && isdigit(code[i+2])) {
if (board.empty()) board = code.substr(i,3);
}
}
return {cpu, memory, board};
}
// Count occurrences across all codes
void count_spares(const vector<string> &codes,
map<string,int> &cpu_cnt,
map<string,int> &mem_cnt,
map<string,int> &board_cnt) {
for (auto &c : codes) {
auto [cpu, mem, brd] = parse_code(c);
if (!cpu.empty()) cpu_cnt[cpu]++;
if (!mem.empty()) mem_cnt[mem]++;
if (!brd.empty()) board_cnt[brd]++;
}
}
// Format map as "key,value;key,value;..."
string format_output(const map<string,int> &cnt) {
string out;
bool first = true;
for (auto &kv : cnt) {
if (!first) out += ";";
out += kv.first + "," + to_string(kv.second);
first = false;
}
return out;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N;
cin >> N;
vector<string> codes(N);
for (int i = 0; i < N; ++i) {
cin >> codes[i];
}
map<string,int> cpu_cnt, mem_cnt, board_cnt;
count_spares(codes, cpu_cnt, mem_cnt, board_cnt);
cout << format_output(cpu_cnt) << "\n";
cout << format_output(mem_cnt) << "\n";
cout << format_output(board_cnt) << "\n";
return 0;
}
Java：[此代码未进行大量数据的测试，仅供参考]
import java.util.*;
public class SpareCounter {
// Parse out first C##, M##, B## segments
public static String[] parseCode(String code) {
String cpu = "", mem = "", board = "";
int L = code.length();
for (int i = 0; i + 2 < L; i++) {
char t = code.charAt(i);
if (t == 'C' && Character.isDigit(code.charAt(i+1)) && Character.isDigit(code.charAt(i+2))) {
if (cpu.isEmpty()) cpu = code.substring(i, i+3);
} else if (t == 'M' && Character.isDigit(code.charAt(i+1)) && Character.isDigit(code.charAt(i+2))) {
if (mem.isEmpty()) mem = code.substring(i, i+3);
} else if (t == 'B' && Character.isDigit(code.charAt(i+1)) && Character.isDigit(code.charAt(i+2))) {
if (board.isEmpty()) board = code.substring(i, i+3);
}
}
return new String[]{cpu, mem, board};
}
// Count occurrences across all codes
public static void countSpares(List<String> codes,
Map<String,Integer> cpuCnt,
Map<String,Integer> memCnt,
Map<String,Integer> boardCnt) {
for (String c : codes) {
String[] parts = parseCode(c);
if (!parts[0].isEmpty()) cpuCnt.put(parts[0], cpuCnt.getOrDefault(parts[0],0) + 1);
if (!parts[1].isEmpty()) memCnt.put(parts[1], memCnt.getOrDefault(parts[1],0) + 1);
if (!parts[2].isEmpty()) boardCnt.put(parts[2], boardCnt.getOrDefault(parts[2],0) + 1);
}
}
// Format map as "key,value;key,value;..."
public static String formatOutput(Map<String,Integer> cnt) {
StringBuilder sb = new StringBuilder();
List<String> keys = new ArrayList<>(cnt.keySet());
Collections.sort(keys);
for (int i = 0; i < keys.size(); i++) {
if (i > 0) sb.append(";");
String k = keys.get(i);
sb.append(k).append(",").append(cnt.get(k));
}
return sb.toString();
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
List<String> codes = new ArrayList<>();
for (int i = 0; i < N; i++) {
codes.add(sc.next());
}
Map<String,Integer> cpuCnt = new HashMap<>();
Map<String,Integer> memCnt = new HashMap<>();
Map<String,Integer> boardCnt = new HashMap<>();
countSpares(codes, cpuCnt, memCnt, boardCnt);
System.out.println(formatOutput(new TreeMap<>(cpuCnt)));
System.out.println(formatOutput(new TreeMap<>(memCnt)));
System.out.println(formatOutput(new TreeMap<>(boardCnt)));
sc.close();
}
}
Python：[此代码未进行大量数据的测试，仅供参考]
# 解析物料编码
def parse_code(code):
cpu = None
memory = None
board = None
for i in range(len(code) - 2):
if code[i] == 'C'and code[i + 1].isdigit() and code[i + 2].isdigit():
if cpu isNone:
cpu = code[i:i + 3]
elif code[i] == 'M'and code[i + 1].isdigit() and code[i + 2].isdigit():
if memory isNone:
memory = code[i:i + 3]
elif code[i] == 'B'and code[i + 1].isdigit() and code[i + 2].isdigit():
if board isNone:
board = code[i:i + 3]
return cpu, memory, board
# 统计备件数量
def count_spares(codes):
cpu_count = {}
memory_count = {}
board_count = {}
for code in codes:
cpu, memory, board = parse_code(code)
if cpu:
cpu_count[cpu] = cpu_count.get(cpu, 0) + 1
if memory:
memory_count[memory] = memory_count.get(memory, 0) + 1
if board:
board_count[board] = board_count.get(board, 0) + 1
return cpu_count, memory_count, board_count
# 格式化输出
def format_output(count_dict):
items = sorted(count_dict.items())
return';'.join([f'{key},{value}'for key, value in items])
N = int(input())
codes = input().split()
cpu_count, memory_count, board_count = count_spares(codes)
print(format_output(cpu_count))
print(format_output(memory_c
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