原文：LEAK: Sonnet 4.7 release will be accelerated during christmas after GPT 5.2 smashes all benchmarks

目前仍没有API版3 flash出现，下面全部数据基于在竞技场/官方渠道手动输入进行。部分为单轮，没有完整平均分数，仅体现趋势，具体数值需要等待最终API版本。 [image] 上一个检查点skyhawk还和2.5pro一样，现在已经完全超越2.5pro了 先前测试 100K-召回率100%?! Gemini 3 flash临时捞针成绩！ - 开发调优 - LINUX DO 可以认为，G…

序号 问题 / 初始素材 状态 推理过程 & 结果截图 1 [image]white_check_mark[PixPin_2025-12-14_15-03-46][PixPin_2025-12-14_15-03-59] 2 [PixPin_2025-04-18_11-39-55] white_check_mark[PixPin_2025-12-14_15…

梭哈！！

class Result {
    int[] dp0;
    int[] dp1;
    int size;

    Result(int[] dp0, int[] dp1, int size) {
        this.dp0 = dp0;
        this.dp1 = dp1;
        this.size = size;
    }
}

class Solution {
    public int maxProfit(int n, int[] present, int[] future, int[][] hierarchy, int budget) {
        List<Integer>[] g = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            g[i] = new ArrayList<>();
        }
        for (int[] e : hierarchy) {
            g[e[0] - 1].add(e[1] - 1);
        }

        return dfs(0, present, future, g, budget).dp0[budget];
    }

    private Result dfs(int u, int[] present, int[] future, List<Integer>[] g, int budget) {
        int cost = present[u];
        int dCost = present[u] / 2;
        // dp[u][state][budget]
        // state = 0: 不购买父节点, state = 1: 必须购买父节点
        int[] dp0 = new int[budget + 1];
        int[] dp1 = new int[budget + 1];

        // subProfit[state][budget]
        // state = 0: 优惠不可用, state = 1: 优惠可用
        int[] subProfit0 = new int[budget + 1];
        int[] subProfit1 = new int[budget + 1];
        int uSize = cost;

        for (int v : g[u]) {
            Result childResult = dfs(v, present, future, g, budget);
            uSize += childResult.size;

            for (int i = budget; i >= 0; i--) {
                for (int sub = 0; sub <= Math.min(childResult.size, i); sub++) {
                    if (i - sub >= 0) {
                        subProfit0[i] = Math.max(subProfit0[i], subProfit0[i - sub] + childResult.dp0[sub]);
                        subProfit1[i] = Math.max(subProfit1[i], subProfit1[i - sub] + childResult.dp1[sub]);
                    }
                }
            }
        }

        for (int i = 0; i <= budget; i++) {
            dp0[i] = subProfit0[i];
            dp1[i] = subProfit0[i];
            if (i >= dCost) {
                dp1[i] = Math.max(subProfit0[i], subProfit1[i - dCost] + future[u] - dCost);
            }
            if (i >= cost) {
                dp0[i] = Math.max(subProfit0[i], subProfit1[i - cost] + future[u] - cost);
            }
        }

        return new Result(dp0, dp1, uSize);
    }
}

作者：力扣官方题解
链接：https://leetcode.cn/problems/maximum-profit-from-trading-stocks-with-discounts/solutions/3851959/zhe-kou-jie-jiao-yi-gu-piao-de-zui-da-li-c7l9/
来源：力扣（LeetCode）
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