大整数哈希

求解代码

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 mod 2 64 \text{mod}~2^{64} mod 264结果完全一致。