题解#1

这是一道经典的“最接近的两个数”问题。核心思路是：排序后，最小差值一定出现在相邻元素之间。
思路
1.
将数组排序。
2.
遍历相邻元素，计算差值$a[i+1] - a[i]$，记录最小值。
3.
若出现更小的差值，更新答案；若差值相等，由于数组已排序，先遇到的相邻对分数一定更小，所以无需额外处理（自然保留第一组）。
4.
输出最小差值和对应的两个数。
时间复杂度$O(n \log n)$，对于$n \le 10^6$完全够用。

C++代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];

    sort(a.begin(), a.end());

    long long minDiff = LLONG_MAX;
    long long x = 0, y = 0;
    for (int i = 0; i < n - 1; i++) {
        long long diff = a[i + 1] - a[i];
        if (diff < minDiff) {
            minDiff = diff;
            x = a[i];
            y = a[i + 1];
        }
    }

    cout << minDiff << "\n" << x << " " << y << "\n";
    return 0;
}

$python$代码：

import sys

def main():
    data = sys.stdin.buffer.read().split()
    n = int(data[0])
    a = sorted(map(int, data[1:n+1]))

    min_diff = float('inf')
    x = y = 0
    for i in range(n - 1):
        diff = a[i+1] - a[i]
        if diff < min_diff:
            min_diff = diff
            x, y = a[i], a[i+1]

    print(min_diff)
    print(x, y)

main()