这里我们会看到一个有趣的问题。我们有一个包含 n 个元素的数组。我们必须执行一个查询 q，如下所示：
q(start, end) 表示从开始到结束，数字“p”出现的次数恰好是“p”次。 p>
因此，如果数组类似于：{1, 5, 2, 3, 1, 3, 5, 7, 3, 9, 8}，并且查询为 -
q(1 , 8) - 这里 1 出现一次，3 出现 3 次。所以答案是 2
q(0, 2) - 这里 1 出现一次。所以答案是 1
算法query(s, e) -
begin get the elements and count the frequency of each element ‘e’ into one map count := count + 1 for each key-value pair p, do if p.key = p.value, then count := count + 1 done return count;end
示例#include <iostream>#include <map>using namespace std;int query(int start, int end, int arr[]) { map<int, int> freq; for (int i = start; i <= end; i++) //get element and store frequency freq[arr[i]]++; int count = 0; for (auto x : freq) if (x.first == x.second) //when the frequencies are same, increase count count++; return count;}int main() { int a[] = {1, 5, 2, 3, 1, 3, 5, 7, 3, 9, 8}; int n = sizeof(a) / sizeof(a[0]); int queries[][3] = {{ 0, 1 }, { 1, 8 }, { 0, 2 }, { 1, 6 }, { 3, 5 }, { 7, 9 } }; int query_count = sizeof(queries) / sizeof(queries[0]); for (int i = 0; i < query_count; i++) { int start = queries[i][0]; int end = queries[i][1]; cout << "answer for query " << (i + 1) << " = " << query(start, end, a) << endl; }}
输出answer for query 1 = 1answer for query 2 = 2answer for query 3 = 1answer for query 4 = 1answer for query 5 = 1answer for query 6 = 0
以上就是在c程序中，将数组范围查询与频率相同的元素进行翻译的详细内容。