C++ を使用して、インデックス範囲内の指定された配列のビットごとの OR 演算をクエリします。

Input: arr[] = {1, 3, 1, 2, 3, 4}, q[] = {{0, 1}, {3, 5}}
Output:
3
7
1 OR 3 = 3
2 OR 3 OR 4 = 7
Input: arr[] = {1, 2, 3, 4, 5}, q[] = {{0, 4}, {1, 3}}
Output:
7
7

#include <bits/stdc++.h>
using namespace std;
int main() {
   int arr[] = { 7, 5, 3, 5, 2, 3 };
   int n = sizeof(arr) / sizeof(int); // size of our array
   int queries[][2] = { { 1, 3 }, { 4, 5 } }; // given queries
   int q = sizeof(queries) / sizeof(queries[0]); // number of queries
   for(int i = 0; i < q; i++) { // traversing through all the queries
      long ans = 0;
      for(int j = queries[i][0]; j <= queries[i][1]; j++) // traversing through the range
         ans |= arr[j]; // calculating the answer
      cout << ans << "\n";
   }
   return 0;
}

#include <bits/stdc++.h>

using namespace std;
#define bitt 32
#define MAX (int)10e5

int prefixbits[bitt][MAX];
void bitcount(int *arr, int n) { // making prefix counts
   for (int j = 31; j >= 0; j--) {
      prefixbits[j][0] = ((arr[0] >> j) & 1);
      for (int i = 1; i < n; i++) {
         prefixbits[j][i] = arr[i] & (1LL << j);
         prefixbits[j][i] += prefixbits[j][i - 1];
      }
   }
   return;
}
int check(int l, int r) { // calculating the answer
   long ans = 0; // to avoid overflow we are taking ans as long
   for (int i = 0; i < 32; i++) {
      int x;
      if (l == 0)
         x = prefixbits[i][r];
      else
         x = prefixbits[i][r] - prefixbits[i][l - 1];
      if (x != 0)
         ans = (ans | (1LL << i));
   }
   return ans;
}
int main() {
   int arr[] = {7, 5, 3, 5, 2, 3};
   int n = sizeof(arr) / sizeof(int); // size of our array
   bitcount(arr, n);
   int queries[][2] = {{1, 3}, {4, 5}}; // given queries
   int q = sizeof(queries) / sizeof(queries[0]); // number of queries
   for (int i = 0; i < q; i++) {
      cout << check(queries[i][0], queries[i][1]) << "\n";
   }
   return 0;
}