给定一个整数K和一个由{L，R}类型的N个查询组成的数组Q [] [] ，每个查询的任务是打印范围[L，R]中不包含整数的数字计数十进制或八进制表示形式的数字K。
例子：
Input: K = 7, Q[][] = {{1, 15}}
Output: 13
Explanation:
Only number 7 and 15 contains digit K in its octal or decimal representations.
Octal representation of number 7 is 7.
Octal representation of number 15 is 17.
Input: K = 8, Q[][] = {{1, 10}}
Output: 9
Explanation:
Only number 8 contains digit K in its decimal representations.
Octal representation of number 8 is 10.
天真的方法：这个想法是遍历每个查询的范围[L，R] ，并找到那些其十进制和八进制表示不包含数字K的数字。打印这些数字的计数。
时间复杂度： O(N 2 *(log 10 (N)+ log 8 (N)))
辅助空间： O(N)
高效的方法：为了优化上述方法，我们的想法是使用Prefix Sum技术预先计算所有此类数字的计数。请按照以下步骤解决问题：
- 初始化一个数组，例如pref [] ，以存储范围为[0，i]的数字计数，该数字的八进制或十进制表示形式包含数字K。
- 现在遍历范围[0，1e6]并执行以下步骤：
- 如果数字以八进制表示形式或十进制表示形式包含数字K ，则将pref [i]更新为pref [i – 1] +1 。
- 否则，将pref [i]更新为pref [i – 1] 。
- 遍历给定的查询并将每个查询的计数打印为
Q[i][1] – Q[i][0] + 1 – (pref[Q[i][1]] – pref[Q[i][0] – 1])
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if the given digit
// 'K' is present in the decimal and octal
// representations of num or not
bool contains(int num, int K, int base)
{
// Stores if the digit exists
// or not
bool isThere = 0;
// Iterate till nums is non-zero
while (num) {
// Find the remainder
int remainder = num % base;
// If the remainder is K
if (remainder == K) {
isThere = 1;
}
num /= base;
}
return isThere;
}
// Function to count the numbers in the
// range [1, N] such that it doesn't
// contain the digit 'K' in its decimal
// and octal representation
void count(int n, int k,
vector > v)
{
// Stores count of numbers in the
// range [0, i] that contains the
// digit 'K' in its octal or
// decimal representation
int pref[1000005] = { 0 };
// Traverse the range [0, 1e6 + 5]
for (int i = 1; i < 1e6 + 5; i++) {
// Check if i contains the digit
// 'K' in its decimal or
// octal representation
bool present
= contains(i, k, 10)
|| contains(i, k, 8);
// Update pref[i]
pref[i] += pref[i - 1] + present;
}
// Print the answer of queries
for (int i = 0; i < n; ++i) {
cout << v[i][1] - v[i][0] + 1
- (pref[v[i][1]]
- pref[v[i][0] - 1])
<< ' ';
}
}
// Driver Code
int main()
{
int K = 7;
vector > Q = { { 2, 5 },
{ 1, 15 } };
int N = Q.size();
// Function Call
count(N, K, Q);
} Java
// Java program for the above approach
import java.util.*;
public class GFG
{
// Function to check if the given digit
// 'K' is present in the decimal and octal
// representations of num or not
static boolean contains(int num, int K, int Base)
{
// Stores if the digit exists
// or not
boolean isThere = false;
// Iterate till nums is non-zero
while (num > 0)
{
// Find the remainder
int remainder = num % Base;
// If the remainder is K
if (remainder == K)
{
isThere = true;
}
num /= Base;
}
return isThere;
}
// Function to count the numbers in the
// range [1, N] such that it doesn't
// contain the digit 'K' in its decimal
// and octal representation
static void count(int n, int k,
Vector > v)
{
// Stores count of numbers in the
// range [0, i] that contains the
// digit 'K' in its octal or
// decimal representation
int[] pref = new int[1000005];
// Traverse the range [0, 1e6 + 5]
for (int i = 1; i < 1e6 + 5; i++) {
// Check if i contains the digit
// 'K' in its decimal or
// octal representation
boolean present
= contains(i, k, 10)
|| contains(i, k, 8);
// Update pref[i]
if(present)
{
pref[i] += pref[i - 1] + 1;
}
}
// Print the answer of queries
System.out.print((v.get(0).get(1) - v.get(0).get(0) +
1 - (pref[v.get(0).get(1)] -
pref[v.get(0).get(0) - 1])) + " ");
System.out.print((v.get(1).get(1) - v.get(1).get(0) -
(pref[v.get(1).get(1)] -
pref[v.get(1).get(0) - 1])) + " ");
}
// Driver code
public static void main(String[] args)
{
int K = 7;
Vector > Q = new Vector >();
Q.add(new Vector());
Q.get(0).add(2);
Q.get(0).add(5);
Q.add(new Vector());
Q.get(1).add(1);
Q.get(1).add(15);
int N = Q.size();
// Function Call
count(N, K, Q);
}
}
// This code is contributed by divyeshrabadiya07. Python3
# Python3 program for the above approach
# Function to check if the given digit
# 'K' is present in the decimal and octal
# representations of num or not
def contains(num, K, base):
# Stores if the digit exists
# or not
isThere = 0
# Iterate till nums is non-zero
while (num):
# Find the remainder
remainder = num % base
# If the remainder is K
if (remainder == K):
isThere = 1
num //= base
return isThere
# Function to count the numbers in the
# range [1, N] such that it doesn't
# contain the digit 'K' in its decimal
# and octal representation
def count(n, k, v):
# Stores count of numbers in the
# range [0, i] that contains the
# digit 'K' in its octal or
# decimal representation
pref = [0]*1000005
# Traverse the range [0, 1e6 + 5]
for i in range(1, 10 ** 6 + 5):
# Check if i contains the digit
# 'K' in its decimal or
# octal representation
present = contains(i, k, 10) or contains(i, k, 8)
# Update pref[i]
pref[i] += pref[i - 1] + present
# Prthe answer of queries
for i in range(n):
print(v[i][1] - v[i][0] + 1- (pref[v[i][1]]- pref[v[i][0] - 1]),end=" ")
# Driver Code
if __name__ == '__main__':
K = 7
Q = [ [ 2, 5 ],
[ 1, 15 ] ]
N = len(Q)
# Function Call
count(N, K, Q)
# This code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to check if the given digit
// 'K' is present in the decimal and octal
// representations of num or not
static bool contains(int num, int K, int Base)
{
// Stores if the digit exists
// or not
bool isThere = false;
// Iterate till nums is non-zero
while (num > 0)
{
// Find the remainder
int remainder = num % Base;
// If the remainder is K
if (remainder == K)
{
isThere = true;
}
num /= Base;
}
return isThere;
}
// Function to count the numbers in the
// range [1, N] such that it doesn't
// contain the digit 'K' in its decimal
// and octal representation
static void count(int n, int k,
List > v)
{
// Stores count of numbers in the
// range [0, i] that contains the
// digit 'K' in its octal or
// decimal representation
int[] pref = new int[1000005];
// Traverse the range [0, 1e6 + 5]
for (int i = 1; i < 1e6 + 5; i++) {
// Check if i contains the digit
// 'K' in its decimal or
// octal representation
bool present
= contains(i, k, 10)
|| contains(i, k, 8);
// Update pref[i]
if(present)
{
pref[i] += pref[i - 1] + 1;
}
}
// Print the answer of queries
Console.Write((v[0][1] - v[0][0] + 1 - (pref[v[0][1]] - pref[v[0][0] - 1])) + " ");
Console.Write((v[1][1] - v[1][0] - (pref[v[1][1]] - pref[v[1][0] - 1])) + " ");
}
// Driver code
static void Main()
{
int K = 7;
List > Q = new List>();
Q.Add(new List());
Q[0].Add(2);
Q[0].Add(5);
Q.Add(new List());
Q[1].Add(1);
Q[1].Add(15);
int N = Q.Count;
// Function Call
count(N, K, Q);
}
}
// This code is contributed by divyesh072019. 输出：
4 13时间复杂度： O(N *(log 10 (N)+ log 8 (N)))
辅助空间： O(N)