UVA10909 Lucky Number题解

最新推荐文章于 2020-10-21 09:04:02 发布

海岛Blog

最新推荐文章于 2020-10-21 09:04:02 发布

阅读量652

点赞数

分类专栏： # ICPC-问题与代码 # 转发转载文章标签： Lucky Number UVA10909

转发转载同时被 2 个专栏收录

64 篇文章

订阅专栏

ICPC-问题与代码

27 篇文章

订阅专栏

本文介绍了解决UVa在线评测系统中10909号问题“Lucky Number”的一种方法。该问题通过构建平衡二叉搜索树并使用迭代消除过程来生成幸运数序列。文章提供了完整的C++代码实现，并详细解释了关键步骤。

原文链接：http://www.algorithmist.com/index.php/User:Sweepline/UVa_10909.cpp

AC的C++语言程序：

/*
 *  Solution for problem 10909 'Lucky Number'.
 *
 *  The code uses a (simple) binary search tree to build the list of
 *  lucky numbers.
 */
#include <cstdio>
#include <cstring>
 
#define MAXN 670000	/* max. number of nodes in the tree */
 
/* Tree's housekeeping...*/
int left[MAXN], right[MAXN], parent[MAXN], key[MAXN], count[MAXN], N, root;
 
char lucky[2010000];
 
/* Returns the index of the k-th smallest element in the tree. */
int find(int k)
{
	for (int x = root;;)
		if (k < count[left[x]])
			x = left[x];
		else if (k == count[left[x]])
			return x;
		else
			k -= count[left[x]] + 1, x = right[x];
}
 
/*
 *  Removes the element with index 'x' from the tree.
 *  Implemented algorithm ensures that the height of the tree does not increase.
 */
void rm(int x)
{
	int y;
 
	if (left[x] != 0 && right[x] != 0) {
		if (count[right[x]] >= count[left[x]])
			for (y = right[x]; left[y] != 0; y = left[y]);
		else
			for (y = left[x]; right[y] != 0; y = right[y]);
		key[x] = key[y];
		x = y;
	}
 
	if (left[x] == 0 && right[x] == 0) {
		if (left[parent[x]] == x)
			left[parent[x]] = 0;
		else
			right[parent[x]] = 0;
	} else {
		y = (left[x] == 0) ? right[x] : left[x];
 
		if (parent[x] == 0) {
			parent[root = y] = 0;
			return;
		}
 
		if (left[parent[x]] == x)
			left[parent[x]] = y;
		else
			right[parent[x]] = y;
		parent[y] = parent[x];
	}
 
	for (x = parent[x]; x != 0; x = parent[x])
		count[x]--;
}
 
/* Constructs a balanced tree with b-a+1 elements; returns index of its root.
   The tree's nodes will get consecutive indices in their order in the tree. */
int build(int a, int b)
{
	if (a > b) return 0;
	if (a == b) { N++; left[N] = right[N] = 0; count[N] = 1; return N; }
 
	int c=(a+b)/2, t = build(a, c-1);
	left[++N] = t; t = N; right[t] = build(c+1, b);
	count[t] = count[left[t]] + count[right[t]] + 1;
	parent[left[t]] = parent[right[t]] = t;
	return t;
}
 
void mark(int x)
{
	for (; x; x = right[x])
		lucky[key[x]] = 1, mark(left[x]);
}
 
/* Constructs the list of lucky numbers */
void make()
{
	int i, j, k;
 
	/* First off, initialize the tree... */
 
	N = count[0] = 0;
	parent[root = build(0, 666667)] = 0;
 
	/*
	 *  As an optimization, we start with the tree, containing all numbers
	 *  of form 6k+1 and 6k+3 in the range of interest.
	 *  These are the numbers, which remain after the first two elimination
	 *  rounds.
	 */
	for (i = 1, j = 1; i <= 666700; j += 6)
		key[i++] = j, key[i++] = j+2;
 
	/* Now just simulation... */
	for (k = 2; k < count[root]; k++) {
		j = key[find(k)]-1;
		if (j >= count[root]) break;
 
		for (i = j; i < count[root]; i += j)
			rm(find(i));
	}
 
	/* Finally, mark the remaining numbers in the boolean array lucky[] */
	memset(lucky, 0, sizeof(lucky));
	mark(root);
}
 
int main()
{
	int a, n;
 
	for (make(); scanf("%d", &n) == 1;) {
		a = 0;
		if (n >= 1 && (n & 1) == 0) {
			for (a = n/2; a > 0 && !lucky[a]; a--);
			for (; a > 0; a -= 2)
				if (lucky[a] && lucky[n-a]) break;
		}
 
		if (a <= 0)
			printf("%d is not the sum of two luckies!\n", n);
		else
			printf("%d is the sum of %d and %d.\n", n, a, n-a);
	}
 
	return 0;
}