POJ 3370 Halloween treats 鸽巢原理 解题

Halloween treats

和POJ2356差点儿相同。

事实上这种数列能够有非常多，也能够有不连续的，只是利用鸽巢原理就是方便找到了连续的数列。并且有这种数列也必然能够找到。

#include <cstdio>
#include <cstdlib>
#include <xutility>

int main()
{
	int c, n;
	while (scanf("%d %d", &c, &n) && c)
	{
		int *neighbours = (int *) malloc(sizeof(int) * n);
		int *sumMod = (int *) malloc(sizeof(int) * (n+1));
		int *iiMap = (int *) malloc(sizeof(int) * c);
		std::fill(iiMap, iiMap+c, -1);

		sumMod[0] = 0;
		int L = -1, R = -1;

		for (int i = 0; i < n; i++) scanf("%d", &neighbours[i]);
		
		for (int i = 0; i < n; i++)
		{
			sumMod[i+1] = (sumMod[i] + neighbours[i]) % c;

			if (sumMod[i+1] == 0)
			{
				L = 1, R = ++i;//下标从1起
				break;
			}

			if (iiMap[sumMod[i+1]] != -1)
			{
				L = iiMap[sumMod[i+1]] + 2, R = ++i; //下标从1起
				break;
			}

			iiMap[sumMod[i+1]] = i;
		}

		if (R != -1)
		{
			for (int i = L; i < R; i++)
			{
				printf("%d ", i);
			}
			printf("%d\n", R);
		}
		else puts("no sweets");
		
		free(neighbours), free(iiMap), free(sumMod);
	}
	return 0;
}