想在明年毕业前买部二手的iphone，求教懂行的大佬，双11，元旦和春节，在哪个时段买价格最好，我意思是相对降幅最大呢
习惯和mac联动，不考虑安卓了，打算买美版iphone15pro无锁版，佬友还有什么建议吗？

API Error: 400 {“error”:{“message”:“Error from provider(xxx,deepseek-v3: 400):
{"error":{"message":"Requested token count exceeds the model’s maximum context length of 163840 tokens.
You requested a total of 167061 tokens: 159061 tokens from the input messages and 8000 tokens for the
completion. Please reduce the number of tokens in the input messages or the completion to fit within the
limit.","type":"bad_response_status_code","param":"","code":"bad_response_status_code"}}Error:
Error from provider(xxx,deepseek-v3: 400): {"error":{"message":"Requested token count exceeds the
model’s maximum context length of 163840 tokens. You requested a total of 167061 tokens: 159061 tokens from
the input messages and 8000 tokens for the completion. Please reduce the number of tokens in the input
messages or the completion to fit within the
limit.","type":"bad_response_status_code","param":"","code":"bad_response_status_code"}}\n
at nt (AppData\Local\nvm\v24.8.0\node_modules\@musistudio\claude-code-router\dist\
cli.js:79940:11)\n    at h0 (AppData\Local\nvm\v24.8.0\node_modules\@musistudio\clau
de-code-router\dist\cli.js:79998:11)\n    at process.processTicksAndRejections
(node:internal/process/task_queues:105:5)\n    at async l0
(AppData\Local\nvm\v24.8.0\node_modules\@musistudio\claude-code-router\dist\cli.js
:79965:96)”,“type”:“api_error”,“code”:“provider_response_error”}}

                 Context left until auto-compact: 11%


“Router”: {
“default”: “xxx,deepseek-v3”,
“background”: “xxx,deepseek-v3”,
“think”: “xxx,deepseek-v3”,
“longContext”: “xxx,deepseek-v3”,
“longContextThreshold”: 115000,
“autoCompactThreshold”: 0.70,
“webSearch”: “”,
“image”: “”
},

现在我的cc跑起来简简单单就爆掉了，修改了config的上下文限制也无济于事，佬友们有没有懂的，我看cc的官方文档里面只有对最大输出的限制，没有调整这个上下文窗口的配置，这个api的上下文窗口是128k

从评测来看，感觉荣耀是个全能水桶机，oppo某些方便比如系统和质感相对优秀
【标准旗舰妥妥够用？小米17，OPPO Find X9，vivo X300和荣耀Magic8户外体验-哔哩哔哩】 https://b23.tv/8IDbKui
【findx9pro深度体验，下次不要画蛇添足！-哔哩哔哩】 https://b23.tv/t1XLypn

是因为copilot完全没竞争力吗？我平常开发基本只用tab补全，copilot还能国内卡付款，价格也便宜，对我来说是完全够用了。

// ==UserScript==
// @name         KYX批量抽奖助手
// @namespace    https://api.kkyyxx.xyz
// @version      1.0.0
// @description  批量抽奖+自动购买
// @match        https://quota.kyx03.de/*
// @author       kkkyyx
// @icon https://linux.do/user_avatar/linux.do/kkkyyx/288/1034473_2.png
// @grant        none
// @run-at       document-end
// ==/UserScript==

(function() {
    'use strict';

    console.log('🎰 KYX脚本开始加载...');

    const API = {
        SPIN: 'https://quota.kyx03.de/api/slot/spin',
        BUY: 'https://quota.kyx03.de/api/slot/buy-spins',
        RECORDS: 'https://quota.kyx03.de/api/slot/records'
    };

    let running = false;
    let panelOpen = false;
    let mode = 'free';
    
    let state = {
        quota: 0,
        spins: 0,
        freeSpins: 0,
        bought: 0,
        maxBuy: 5,
        price: 20000000
    };
    
    let stats = {
        spins: 0,
        win: 0,
        spent: 0
    };

    const isMobile = /Android|iPhone|iPad/i.test(navigator.userAgent) || window.innerWidth < 768;

    function $(id) {
        return document.getElementById(id);
    }

    // 创建悬浮球
    function createBall() {
        console.log('创建悬浮球...');
        
        const ball = document.createElement('div');
        ball.id = 'kyx-ball';
        const size = isMobile ? 60 : 70;
        
        ball.style.position = 'fixed';
        ball.style.bottom = (isMobile ? 20 : 30) + 'px';
        ball.style.right = (isMobile ? 20 : 30) + 'px';
        ball.style.width = size + 'px';
        ball.style.height = size + 'px';
        ball.style.background = 'linear-gradient(135deg, #667eea, #764ba2)';
        ball.style.borderRadius = '50%';
        ball.style.display = 'flex';
        ball.style.alignItems = 'center';
        ball.style.justifyContent = 'center';
        ball.style.cursor = 'pointer';
        ball.style.zIndex = '999998';
        ball.style.boxShadow = '0 4px 20px rgba(102,126,234,0.5)';
        ball.style.transition = 'all 0.3s';
        ball.style.border = '3px solid white';
        
        ball.innerHTML = '<div style="font-size: 40px;">🎰</div>';
        
        ball.addEventListener('click', function() {
            console.log('悬浮球被点击');
            togglePanel();
        });
        
        ball.addEventListener('mouseenter', function() {
            ball.style.transform = 'scale(1.1)';
        });
        
        ball.addEventListener('mouseleave', function() {
            ball.style.transform = 'scale(1)';
        });
        
        document.body.appendChild(ball);
        console.log('✅ 悬浮球创建成功');
    }

    // 创建面板
    function createPanel() {
        console.log('创建面板...');
        
        const panel = document.createElement('div');
        panel.id = 'kyx-panel';
        
        panel.style.position = 'fixed';
        panel.style.top = '50%';
        panel.style.left = '50%';
        panel.style.transform = 'translate(-50%, -50%) scale(0.95)';
        panel.style.width = isMobile ? '90%' : '400px';
        panel.style.maxWidth = '450px';
        panel.style.background = 'white';
        panel.style.borderRadius = '20px';
        panel.style.boxShadow = '0 10px 40px rgba(0,0,0,0.3)';
        panel.style.zIndex = '999999';
        panel.style.display = 'none';
        panel.style.opacity = '0';
        panel.style.transition = 'all 0.3s';
        panel.style.maxHeight = '90vh';
        panel.style.overflow = 'auto';

        panel.innerHTML = `
            <div style="background: linear-gradient(135deg, #667eea, #764ba2); padding: 20px; border-radius: 20px 20px 0 0; color: white; position: sticky; top: 0; z-index: 10;">
                <div style="display: flex; justify-content: space-between; align-items: center;">
                    <h3 style="margin: 0; font-size: 18px;">🎰 批量抽奖助手</h3>
                    <button id="kyx-close" style="background: rgba(255,255,255,0.2); border: none; color: wh

1526. 形成目标数组的子数组最少增加次数
思路
第一反应，贪心+分治就行，这真的是困难题吗？每次贪心区间最小值，然后分段递归结果。
代码
    private int[] target;
    public int minNumberOperations(int[] target) {
        this.target = target;
        return calc(0, 0, target.length - 1);
    }

    public int calc(int low, int left, int right) {
        if (left == right) {
            return target[left] - low;
        }
        ArrayList<int[]> span = new ArrayList<>();
        int min = target[left];
        int last = left;
        for (int i = left + 1; i <= right; i++) {
            if (target[i] < min) {
                min = target[i];
                span.clear();
                span.add(new int[]{left, i - 1});
                last = i;
            } else if (target[i] == min) {
                if (i > last + 1) {
                    span.add(new int[]{last + 1, i - 1});
                }
                last = i;
            }
        }
        if (last <= right - 1) {
            span.add(new int[]{last + 1, right});
        }
        int ans = min - low;
        for (int[] ints : span) {
            ans += calc(min, ints[0], ints[1]);
        }
        return ans;
    }

时间复杂度：O(n^2)
空间复杂度：O(n)
优化思路
但是！但是！写完后发现时间复杂度最差情况下会退化到O(n^2)，看范围1 <= target.length <= 10^5大概率会超时。
我需要优化分段方法，或者换个思路。那么我能不能一次分段呢？
如果把数组看出一个线段，那么波峰波谷之间的差值就是需要操作的次数，中间部分无需关注，这样就能一次分段了。只需要找到所有的波谷分段，分段过程中记录波峰，然后每次进行削峰操作。
代码
    public int minNumberOperations(int[] target) {
        if (target.length == 1) return target[0];
        ArrayList<Integer> list = new ArrayList<>();
        int last = 0;
        list.add(0);
        for (int num : target) {
            if (num != last) {
                list.add(num);
                last = num;
            }
        }
        if (last != 0) list.add(0);
        int top = 0;
        int ans = 0;
        int[] lastBottom = null;
        for (int i = 1; i < list.size() - 1; i++) {
            int v = list.get(i);
            if (v < list.get(i - 1) && v < list.get(i + 1)) {
                if (lastBottom == null) {
                    lastBottom = new int[]{v, top};
                } else {
                    if (v >= lastBottom[0]) {
                        ans += top - lastBottom[0];
                        lastBottom[0] = v;
                    } else {
                        ans += top - v;
                    }
                }
                top = 0;
            } else if (v > top) {
                top = v;
            }
        }
        ans += top;
        if (lastBottom != null) {
            ans += lastBottom[1] - lastBottom[0];
        }
        return ans;
    }

时间复杂度：O(n)
空间复杂度：O(n)
优化思路
PS.一开始想求出分段后依次削峰的，后来发现可以边分边削。
过了，但是！居然是较慢的那一批，已经是O(n)了呀。。。可能使用List影响了效率，另外在考虑削峰的过程中，发现似乎可以根据前缀直接得到是否需要削峰、要削多少，等下午下班再考虑优化下(如果我记得并且懒癌没有发作的话)。

求大佬们推荐
之前用过88code，三天炸两天，客服装死，直接pass

我一个一个的试过了，都能注册。有需要的可以去注册个。

测试时间：
2025年10月30日12点07分

666开头
666037.xyz
666053.xyz
666173.xyz
666305.xyz
666351.xyz
666507.xyz
666或888结尾
017666.xyz
031666.xyz
032666.xyz
037666.xyz
038666.xyz
072666.xyz
073888.xyz
073666.xyz
075666.xyz
059666.xyz
083666.xyz
093666.xyz
079666.xyz
207666.xyz
290666.xyz
308666.xyz
381666.xyz
501666.xyz
590666.xyz
705666.xyz
701666.xyz
702666.xyz
703666.xyz
710666.xyz
721666.xyz
751666.xyz
791666.xyz
803666.xyz
805666.xyz
902666.xyz

原帖子是这个：https://linux.do/t/topic/888279
注册之后，邀请码接龙。
然后今天收到了举报，评论被移除的消息！
我不明白，这是为什么。。。