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大整数哈希

在这里插入图片描述

求解代码

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public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer in = new StringTokenizer(br.readLine());
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

        HashMap<Long, Long> hmap = new HashMap<>();

        long total = 0;
        int n = Integer.parseInt(in.nextToken());

        for (int i = 1; i <= n; i++) {

            in = new StringTokenizer(br.readLine());
            long x = Long.parseUnsignedLong(in.nextToken());
            long y = Long.parseUnsignedLong(in.nextToken());

            long ans = hmap.getOrDefault(x, 0L);

            total += (long) i * ans;

            hmap.put(x, y);
        }

        out.println(Long.toUnsignedString(total));

        out.flush();
        out.close();
        br.close();
    }

小贴士

  • Long.parseUnsignedLong()专门解析 64 位无符号整数。
  • Long.toString()是把最高位当作符号位,输出带正负的有符号数,而Long.toUnsignedString()不会修改内存中的数据,只是换一种规则解读二进制,把所有位都当作数值位,直接解析为无符号十进制数,和题目要求的 $\text{mod}~2^{64}$结果完全一致。
最后更新于 2026-04-05 17:35:33
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