题解 | #分苹果#

using System;
using System.Collections;
using System.Collections.Generic;
public class Program {
    public static void Main() {
        string line;
        while ((line = System.Console.ReadLine ()) != null) { // 注意 while 处理多个 case
            int n = int.Parse(line);
            int[] a = new int[n];
            line = Console.ReadLine();
            string[] tokens = line.Split(' ');
            for(int i = 0; i < n; i++) a[i] = int.Parse(tokens[i]);
            int sum = 0;
            foreach(int i in a) sum += i;
            if(((float)sum / n) % 1 != 0) Console.WriteLine(-1);
            else{
                int avg = sum / n;
                int remain = avg % 2;
                int count = 0;
                foreach(int ai in a){
                    if(ai % 2 != remain){
                        count = -1;
                        break;
                    }
                    else{
                        if(ai > avg){
                            count += (ai - avg) / 2;
                        }
                    }
                }
                Console.WriteLine(count);
            }
        }
    }
}

复杂度O(N)。先计算苹果的平均值，平均值不为整数直接输出-1；对每只牛的苹果遍历，若某只牛的苹果和平均值的奇偶不同，令

count=-1；break；若奇偶相同，且苹果数大于平均值，计算每多2个，count++