<!DOCTYPE html> <html> <head> <title>疯狂点击王</title> <style> body { background-color: #f0f0f0; text-align: center; } #button { width: 200px; height: 200px; background-color: #4CAF50; border: none; border-radius: 50%; font-size: 24px; cursor: pointer; margin: 50px auto; } #score { font-size: 30px; margin: 20px; } #time { fo
代码分析与改进 问题分析 1. 连击逻辑问题：当前的连击逻辑`combo = Math.min(score % 5 + 1, 3);`不是按照连续点击次数来计算连击倍数，而是根据总得分来计算，不符合连击的常规逻辑。 2. 用户体验问题：没有给用户明确的连击提示，用户不清楚当前的连击倍数。 改进方案 1. 修正连击逻辑：使用一个变量来记录连续点击的次数，当连续点击次数达到一定值时增加连击倍数。 2. 增加连击提示：在页面上显示当前的连击倍数。 改进后的代码 ```html <!DOCTYPE html> <html> <head> <title>疯狂点击王</title> <style> body { background-color: #f0f0f0; text-align: center; } #button { width: 200px; height: 200px; background-color: #4CAF50; border: none; border-radius: 50%; font-size: 24px; cursor: pointer; margin: 50px auto; } #score { font-size: 30px; margin: 20px; } #time { font-size: 24px; color: #ff0000; } #combo-display { font-size: 20px; color: #0000ff; } /* 新增连击显示样式 */ </style> </head> <body> <div id="score">得分：0</div> <div id="time">剩余时间：60</div> <div id="combo-display">连击倍数：1</div> <!-- 新增连击显示 --> <button id="button" onclick="clickHandler()">点击我！</button> <script> let score = 0; let time = 60; let combo = 1; // 连击倍数 let comboCount = 0; // 连续点击次数 let timer = setInterval(updateTime, 1000); // 点击事件 function clickHandler() { comboCount++; if (comboCount % 5 === 0 && combo < 3) { combo++; } score += 1 * combo; document.getElementById('score').textContent = `得分：${score}`; document.getElementById('combo-display').textContent = `连击倍数：${combo}`; // 更新连击显示 } // 倒计时 function updateTime() { time--; document.getElementById('time').textContent = `剩余时间：${time}`; if (time <= 0) { clearInterval(timer); document.getElementById('button').disabled = true; alert(`游戏结束！最终得分：${score}`); } } </script> </body> </html> ``` 额外需要注意的逻辑点 1. 连击重置逻辑：如果需要在一定时间内没有点击则重置连击，可以在`updateTime`函数中添加逻辑，当一段时间没有点击时重置`comboCount`和`combo`。 2. 边界条件：确保在计算连击倍数时，倍数不会超出最大值，这里设置的最大值是3。 3. 用户交互：可以考虑在游戏结束后提供重新开始的按钮，让用户可以再次进行游戏。 [2025-05-31 13:36:16 | AI写代码神器 | 814点数解答]
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