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 = 1 Answer for Query 2 = 2 Answer for Query 3 = 1 Answer for Query 4 = 1 Answer for Query 5 = 1 Answer for Query 6 = 0
