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.
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：
输入n
输入n个数
判断这n个数中是不是有几个数字之和是n的倍数
思路：
n个数余数分别为 1 ~ n-1 ,相当于有n-1个抽屉，n个物品
分别计算a[1] + a[2] + …… + a[k] 的和然后取余如果为零则直接输出前k个数，否则寻找余数相同的两个数，假设为i， j （i < j),则a[i+1] + . . . . + a[j] 的和一定能被n整除（原理还没想清楚）
AC代码
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 using namespace std; 5 int a[10005]; 6 int mod[10005]; 7 int mark[10005]; 8 9 int main() 10 { 11 int n; 12 bool flag = false; 13 cin >> n; 14 memset(mod, 0, sizeof(mod)); 15 memset(mark, 0, sizeof(mark)); 16 for(int i = 1; i <= n; i++) 17 { 18 cin >> a[i]; 19 mod[i] = (mod[i-1] + a[i]) % n; 20 } 21 22 for(int i = 1; i <= n; i++) 23 { 24 if(mod[i] == 0) 25 { 26 flag = true; 27 cout << i << endl; 28 for(int j = 1; j <= i; j++) 29 cout << a[j] << endl; 30 break; 31 } 32 } 33 34 if(!flag) 35 { 36 for(int i = 1; i <= n; i++) 37 { 38 if(mark[mod[i]] == 0) 39 mark[mod[i]] = i; 40 else 41 { 42 cout << i -mark[mod[i]] << endl; 43 for(int j = mark[mod[i]]+1; j <= i; j++) 44 cout << a[j] << endl; 45 46 break; 47 } 48 } 49 } 50 51 return 0; 52 }
