鸽巢原理

 

鸽巢原理： n+1个鸽子放入n个窝中，至少有一个窝含有两只鸽子 

Find a multiple

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5590		Accepted: 2434		Special Judge

Description

The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000. This numbers are not necessarily different (so it may happen that two or more of them will be equal). Your task is to choose a few of given numbers ( 1 <= few <= N ) so that the sum of chosen numbers is multiple for N (i.e. N * k = (sum of chosen numbers) for some natural number k).

Input

The first line of the input contains the single number N. Each of next N lines contains one number from the given set.

Output

In case your program decides that the target set of numbers can not be found it should print to the output the single number 0. Otherwise it should print the number of the chosen numbers in the first line followed by the chosen numbers themselves (on a separate line each) in arbitrary order.

If there are more than one set of numbers with required properties you should print to the output only one (preferably your favorite) of them.

Sample Input

5
1
2
3
4
1

Sample Output

2
2
3

题意：

一个集合一共有N个数字 ，从中选取任意一个数字集合，使集合中的数字的和是N的倍数

解题思路：

首先，一定存在这样的集合，它是集合中的一串连续数字，且其和为 N的倍数

证明：

设集合中的数字为 a1,a2,a3,a4,……ai，则设：

S1 = a1

S2 = a1+a2

S3 = a1+a2+a3

Sn = a1+a2+a3+……+an

若 Sn % N = 0,则 Sn 即为所求，

若 Sn % N ！= 0 ，Sn % N 的 范围 在 【1，N-1】之间 ， 又一共有 N 个余数 ，根据鸽巢原理，一定有两个余数是相同的，

我们假设为Si 和 Sj，

即有：

Si % N = t

Sj % N = t

那么一定有：

（Si - Sj）% N = 0

所以一定存在符合要求的解，且其集合中的数字是连续的，范围是【ai，aj】。

证毕。

import java.util.Scanner;public class ACM16 {public static void main(String[] args) {Scanner cin = new Scanner(System.in);int num = cin.nextInt();int[] arr = new int[num], sum = new int[num], b = new int[num];for (int i = 0; i < num; i++) {arr[i] = cin.nextInt();b[i] = -1;}sum[0] = arr[0] % num;b[sum[0]] = 0;int temp = 0, temp1 = 0;for (int i = 1; i < num; i++) {sum[i] = sum[i - 1] + arr[i];sum[i] %= num;if (sum[i] == 0) {temp = -1;temp1 = i;break;}if (b[sum[i]] == -1) {b[sum[i]] = i;} else {temp = b[sum[i]];temp1 = i;break;}}System.out.println(temp1 - temp);for (int i = temp + 1; i <= temp1; i++) {System.out.println(arr[i]);}}
}