UVA10909 Lucky Number题解

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;elsek -= 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]);elsefor (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;elseright[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;elseright[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);elseprintf("%d is the sum of %d and %d.\n", n, a, n-a);}return 0;
}

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