华为笔试题 2022.3.30

article/2025/10/11 5:18:11

1、业务部署芯片

在这里插入图片描述

思路，就硬模拟

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>using namespace std;int main() {int m, n;cin >> m;cin >> n;char arr[n];for (int i = 0; i < n; ++i) {cin >> arr[i];}// int m = 5, n = 6;// char arr[n] = "ABABAA";//以上是输入  m 是芯片数，n是业务数，arr就是一个字符数组，放了一串ABABABABchar temp;int b[m];for (int i = 0; i < m; ++i) {b[i] = 0;}int a1 = 0;int index1 = 1;  //未满状态的第一块芯片编号int index2 = 1;  //全空状态的第一块芯片编号for (int i = 0; i < n - 1; ++i) {temp = arr[i];if (temp == 'A') {if (a1 + 1 == 4) {a1 = 0;index1 = index2;} else {a1++;if (index1 == index2) {index2++;}}} else if (temp == 'B') {if (index1 == index2) {index1++;index2++;} else {index2++;}}}// 判断最后一个if (arr[n - 1] == 'A') {if (index1 > m) {cout << 0 << endl;cout << 0 << endl;} else {cout << index1 << endl;cout << a1 + 1 << endl;}} else {if (index2 > m) {cout << 0 << endl;cout << 0 << endl;} else {cout << index2 << endl;cout << 1 << endl;}}return 0;
}

第二题，标准的二维地图（带障碍物）两点最短距离

在这里插入图片描述

上dfs

#include <algorithm>
#include <iostream>
#include <string>using namespace std;
//全局变量
const int MAX_X = 100;
const int MAX_Y = 100;
int min_step = 10000;
int number = 0;
int my_x, my_y;
int map[MAX_X][MAX_Y];
int book[MAX_X][MAX_Y];
int start_x, start_y;
int dest_x, dest_y;void dfs(int x, int y, int step) {/*up, right, down, left*/int next[4][2] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}};int tx, ty;if (x == dest_x && y == dest_y) {if (step == min_step) {number++;}if (step < min_step) {min_step = step;number = 1;}return;}for (int i = 0; i < 4; i++) {tx = x + next[i][0];ty = y + next[i][1];if (tx > my_x || ty > my_y || tx < 0 || ty < 0)continue;if (map[tx][ty] == 0 && book[tx][ty] == 0) {book[tx][ty] = 1;dfs(tx, ty, step + 1);book[tx][ty] = 0;}}
}int main() {cin >> my_x >> my_y;cin >> start_x >> start_y;cin >> dest_x >> dest_y;int num;cin >> num;//建一张二维表,初始化权为0int x, y;                        //湖泊的位置for (int i = 0; i < num; ++i) {  //高山湖泊用1代表cin >> x >> y;map[x][y] = 1;}book[start_x][start_y] = 1;dfs(start_x, start_y, 0);cout << number << " " << min_step << endl;return 0;
}