题解 | 游游的数值距离

import java.util.Scanner;
import java.util.ArrayList;
import java.util.List;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long minAbs = Long.MAX_VALUE;
        int bestX = -1;
        int bestY = -1;

        List<Integer> xList = new ArrayList<>();
        xList.add(1);
        for (int x = 3; x <= 20; x++) {
            xList.add(x);
        }

        for (int x : xList) {
            long fact = 1;
            for (int i = 2; i <= x; i++) {
                fact *= i;
            }
            long f = fact - 1;
            long currentAbs;
            int currentY;
            if (x == 1) {
                currentAbs = n;
                currentY = 1;
            } else {
                List<Integer> candidates = new ArrayList<>();
                if (f > n) {
                    candidates.add(1);
                } else {
                    int y0 = (int) (n / f);
                    candidates.add(y0 - 1);
                    candidates.add(y0);
                    candidates.add(y0 + 1);
                }
                List<Integer> validCandidates = new ArrayList<>();
                for (int y : candidates) {
                    if (y > 0 && y != 2) {
                        validCandidates.add(y);
                    }
                }
                if (validCandidates.isEmpty()) {
                    validCandidates.add(1);
                    validCandidates.add(3);
                }
                long localMinAbs = Long.MAX_VALUE;
                int localBestY = -1;
                for (int y : validCandidates) {
                    long abs = Math.abs(f * y - n);
                    if (abs < localMinAbs) {
                        localMinAbs = abs;
                        localBestY = y;
                    } else if (abs == localMinAbs) {
                        if (y < localBestY) {
                            localBestY = y;
                        }
                    }
                }
                currentAbs = localMinAbs;
                currentY = localBestY;
            }
            if (currentAbs < minAbs || (currentAbs == minAbs && (x < bestX || (x == bestX &&
                                        currentY < bestY)))) {
                minAbs = currentAbs;
                bestX = x;
                bestY = currentY;
            }
        }
        System.out.println(bestX + " " + bestY);
    }
}

https://www.nowcoder.com/discuss/727521113110073344

思路：

1. 输入处理：读取输入的整数 𝑛。
2. 候选生成：生成可能的 𝑥 值列表，跳过 𝑥=2。
3. 阶乘计算：计算每个 𝑥 的阶乘，并生成对应的候选 𝑦 值。
4. 过滤和选择：过滤无效候选值，选择使表达式绝对值最小的 𝑦。
5. 更新最优解：比较所有候选解，找到并输出最优的 𝑥 和 𝑦。