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杭电网安复试编程Day24

news 2026/5/12 16:14:49

1、十六进制转换

题目描述：输入一个十进制的数，把它转成十六进制。

方法一：利用内置函数

#include<iostream> using namespace std; int n; int main() { cin>>n; cout << hex << n << endl; return 0; }

方法二：手动转换

#include<iostream> using namespace std; int n; void turns(int x) { if (x == 0) { cout << "0"; return; //提前返回 } char hexDigits[] = "0123456789ABCDEF"; char result[100]; // 足够存储十六进制结果 int index = 0; while (x > 0) { int remainder = x % 16; result[index++] = hexDigits[remainder]; x /= 16; } // 逆序输出 for (int i = index - 1; i >= 0; i--) { cout << result[i]; } } int main() { cin >> n; turns(n); return 0; }

2、贪吃蛇

Worm is an old computer game. There are many versions, but all involve maneuvering a "worm" around the screen, trying to avoid running the worm into itself or an obstacle.
We'll simulate a very simplified version here. The game will be played on a 50*50 board, numbered so that the square at the upper left is numbers (1, 1). The worm is initially a string of 20 connected squares. Connected squares are adjacent horizontally or vertically. The worm starts stretched out horizontally in positions (25, 11) through (25, 30), with the head of the worm at (25, 30). The worm can move either East (E), West (W), North (N) or South (S), but will never move back on itself. So, in the initial position, a W move is not possible. Thus the only two squares occupied by the worm that change in any move are its head and tail. Note that the head of the worm can move to the square just vacated by the worm's tail.
You will be given a series of moves and will simulate the moves until either the worm runs into itself, the worm runs off the board, or the worm successfully negotiates its list of moves. In the first two cases you should ignore the remaining moves in the list.
INPUT
There will be multiple problems instances. The input for each problem instance will be on two lines. The first line is an integer n (<100) indicating the number of moves to follow. (A value of n = 0 indicates end of input.) The next line contains n characters (either E, W, N or S), with no spaces separating the letter, indicating the sequence of moves.
OUTPUT
Generate one line of output for each problem instance. The output line should be one of the follow three:
The worm ran into itself on move m.
The worm ran off the board on move m.
The worm successfully made all m moves.
Where m is for you to determine and the first move is move 1.

#include <iostream> #include <string> using namespace std; // 蛇身坐标结构体 struct SnakeSegment { int x, y; }; int main() { int n; // 移动步数 string moves; // 存储移动序列 while (cin >> n && n != 0) { cin >> moves; // 读取移动字符串（无空格） // 初始化蛇身：20节，水平放置，头在 (25,30)，尾在 (25,11) SnakeSegment worm[20]; for (int i = 0; i < 20; ++i) { worm[i].x = 25; worm[i].y = 11 + i; // worm[0]为尾，worm[19]为头 } bool success = true; // 标记是否成功完成所有移动 int moveIdx = 0; // 当前移动序号（从1开始） // 依次执行每个移动指令 for (moveIdx = 0; moveIdx < n; ++moveIdx) { char dir = moves[moveIdx]; // 当前方向 // 计算新头的位置 int newHeadX = worm[19].x; int newHeadY = worm[19].y; switch (dir) { case 'N': newHeadX--; break; // 北 case 'S': newHeadX++; break; // 南 case 'E': newHeadY++; break; // 东 case 'W': newHeadY--; break; // 西 } // 1. 检查是否撞墙（边界 1~50） if (newHeadX < 1 || newHeadX > 50 || newHeadY < 1 || newHeadY > 50) { cout << "The worm ran off the board on move " << moveIdx + 1 << ".\n"; success = false; break; } // 2. 检查是否撞到自己（允许撞到蛇尾，因为蛇尾即将离开） bool selfCollision = false; // 遍历除蛇尾（worm[0]）以外的所有身体 for (int j = 1; j < 20; ++j) { if (newHeadX == worm[j].x && newHeadY == worm[j].y) { selfCollision = true; break; } } if (selfCollision) { cout << "The worm ran into itself on move " << moveIdx + 1 << ".\n"; success = false; break; } // 3. 合法移动：更新蛇身 // 将所有身体向前移动一节（蛇尾消失，新头加入） for (int j = 0; j < 19; ++j) { worm[j].x = worm[j + 1].x; worm[j].y = worm[j + 1].y; } worm[19].x = newHeadX; worm[19].y = newHeadY; } // 如果循环正常结束且未触发任何错误，则成功完成所有移动 if (success) { cout << "The worm successfully made all " << n << " moves.\n"; } } return 0; }

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