标签：hdu2062

Subset sequence

1 1
2 1
2 2
2 3
2 4
3 10

1
1
1 2
2
2 1
2 3 1

#include <stdio.h>
#include <string.h>
#include <math.h>

__int64 an[22] = {0, 1}, m;
int box[22], vis[22];

int find_m_min(int n, int m) {
	int i, k = 0;
	for (i = 1; i <= n; ++i)
		if (!vis[i] && ++k == m) {
			break;
		} 
	vis[i] = 1; return i;
}

int main() {
	freopen("stdin.txt", "r", stdin);
	int i, n, id, j, nn, tmp;
	for (i = 2; i <= 20; ++i)
		an[i] = i * (an[i-1] + 1);
	while (scanf("%d%I64d", &n, &m) == 2) {
		memset(vis, 0, sizeof(vis));
		nn = n;
		while (n-- && m) {
			j = m / (an[n] + 1) + (m % (an[n] + 1) ? 1 : 0);
			printf("%d", find_m_min(nn, j));
			m -= (j - 1) * (an[n] + 1) + 1;
			printf("%c", m ? ' ' : '\n');
		}
	}
	return 0;
}

HDU2062 Subset sequence

标签：hdu2062

原文地址：http://blog.csdn.net/chang_mu/article/details/42869267