题目1371：最小的K个数

时间限制：1 秒内存限制：32 兆题目描述：输入n个整数，找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字，则最小的4个数字是1,2,3,4,。输入：每个测试案例包括2行：第一行为2个整数n，k(1<=n，k<=200000)，表示数组的长度。第二行包含n个整数，表示这n个数，数组中的数的范围是[0,1000 000 000]。输出：对应每个测试案例，输出最小的k个数，并按从小到大顺序打印。样例输入：

8 4

4 5 1 6 2 7 3 8

样例输出：

1 2 3 4

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int N = 200001;int heaps;void precolateDown(int k){int parent = 1;int val = heaps[1];int child = parent << 1;while (child <= k){if (child + 1 <= k && heaps[child + 1] > heaps[child])++child;if (heaps[child] > val){heaps[parent] = heaps[child];parent = child;child = parent << 1;}elsebreak;}heaps[parent] = val;}bool cmp(int a, int b){return a > b;}int main(void){int n, k;while (scanf("%d", &n) != EOF){scanf("%d", &k);for (int i = 1; i <= n; ++i)scanf("%d", &heaps[i]);sort(heaps + 1, heaps + k + 1, cmp);for (int i = k + 1; i <= n; ++i){if (heaps[i] < heaps[1]){swap(heaps[1], heaps[i]);precolateDown(k);}}sort(heaps + 1, heaps + k + 1);for (int i = 1; i <= k; ++i)i == 1 ? printf("%d", heaps[i]) : printf(" %d", heaps[i]);printf("\n");}return 0;}

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