红军补给均衡：相邻站点调拨统计代码附录

站内文件视图直接读取仓库内容，Markdown 使用文档排版渲染，其余文本文件保持原始排版，方便校对训练证据链。

文件类型Markdown

10-cases/s2-jh-06-supply-balance/06-deliverables/appendix-code.md

完整解题档案：complete-solution-dossier.md

Python 主实现

源文件：main.py

实现状态：当前已有可执行实现

import sys


def solve(data: str) -> str:
    tokens = list(map(int, data.split()))
    if not tokens:
        return ""
    it = iter(tokens)
    n = next(it)
    values = [next(it) for _ in range(n)]
    total = sum(values)
    if total % n != 0:
        return "possible=NO"
    target = total // n
    prefix_diff = 0
    moves = 0
    units = 0
    for index in range(n - 1):
        prefix_diff += values[index] - target
        if prefix_diff != 0:
            moves += 1
            units += abs(prefix_diff)
    return "\n".join(
        [
            "possible=YES",
            f"moves={moves}",
            f"units={units}",
        ]
    )


if __name__ == "__main__":
    sys.stdout.write(solve(sys.stdin.read()).strip())
    sys.stdout.write("\n")

C++ 对照实现

源文件：main.cpp

实现状态：当前已有可执行实现

#include <cstdlib>
#include <iostream>
#include <vector>

using namespace std;

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

    int n;
    if (!(cin >> n)) {
        return 0;
    }
    vector<long long> values(n);
    long long total = 0;
    for (int i = 0; i < n; ++i) {
        cin >> values[i];
        total += values[i];
    }
    if (total % n != 0) {
        cout << "possible=NO\n";
        return 0;
    }
    long long target = total / n;
    long long prefix_diff = 0;
    long long units = 0;
    int moves = 0;
    for (int i = 0; i < n - 1; ++i) {
        prefix_diff += values[i] - target;
        if (prefix_diff != 0) {
            ++moves;
            units += llabs(prefix_diff);
        }
    }
    cout << "possible=YES\n";
    cout << "moves=" << moves << "\n";
    cout << "units=" << units << "\n";
    return 0;
}

红军补给均衡：相邻站点调拨统计 代码附录

Python 主实现

C++ 对照实现

红军补给均衡：相邻站点调拨统计代码附录