(组合数学3.3.1.1)POJ 2356 Find a multiple(抽屉原理)

/*
 * POJ_2356.cpp
 *
 *  Created on: 2013年10月8日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>

using namespace std;

const int maxn = 10005;
int a[maxn], s[maxn], mod[maxn];

void print(int s, int t) {
	printf("%d\n", t - s + 1);

	int i;
	for (i = s; i <= t; ++i) {
		printf("%d\n", a[i]);
	}
}

int main() {
	int n;
	scanf("%d", &n);

	memset(a, 0, sizeof(a));
	memset(s, 0, sizeof(s));
	memset(mod, 0, sizeof(mod));

	int i;
	for (i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
		s[i] = s[i-1] + a[i];

		if (s[i] % n == 0) {
			print(1, i);
			break;
		} else if (!mod[s[i] % n]) {
			mod[s[i] % n] = i;
		} else {
			print(mod[s[i] % n] + 1, i);
			break;
		}
	}


}