给定两个正整数N和K ，任务是计算满足以下条件的所有数字：
如果数字是num ，
- num≤N 。
- abs(num – count)≥K ，其中count是最大为num的素数的计数。
例子：
Input: N = 10, K = 3
Output: 5
6, 7, 8, 9 and 10 are the valid numbers. For 6, the difference between 6 and prime numbers upto 6 (2, 3, 5) is 3 i.e. 6 – 3 = 3. For 7, 8, 9 and 10 the differences are 3, 4, 5 and 6 respectively which are ≥ K.
Input: N = 30, K = 13
Output: 10
先决条件：二进制搜索
方法：观察到一个函数，它是质数的数量与质数的差直到该数字，对于一个特定的K而言，是一个单调递增的函数。同样，如果数字X是有效数字，则X +1也是有效数字。
证明 ：
Let the function Ci denotes the count of prime numbers upto number i. Now,
for the number X + 1 the difference is X + 1 – CX + 1 which is greater than
or equal to the difference X – CX for the number X, i.e. (X + 1 – CX + 1) ≥ (X – CX).
Thus, if (X – CX) ≥ S, then (X + 1 – CX + 1) ≥ S.
因此，我们可以使用二进制搜索来找到最小有效数字X，并且从X到N的所有数字都是有效数字。因此，答案将是N – X + 1 。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
const int MAX = 1000001;
// primeUpto[i] denotes count of prime
// numbers upto i
int primeUpto[MAX];
// Function to compute all prime numbers
// and update primeUpto array
void SieveOfEratosthenes()
{
bool isPrime[MAX];
memset(isPrime, 1, sizeof(isPrime));
// 0 and 1 are not primes
isPrime[0] = isPrime[1] = 0;
for (int i = 2; i * i < MAX; i++) {
// If i is prime
if (isPrime[i]) {
// Set all multiples of i as non-prime
for (int j = i * 2; j < MAX; j += i)
isPrime[j] = 0;
}
}
// Compute primeUpto array
for (int i = 1; i < MAX; i++) {
primeUpto[i] = primeUpto[i - 1];
if (isPrime[i])
primeUpto[i]++;
}
}
// Function to return the count
// of valid numbers
int countOfNumbers(int N, int K)
{
// Compute primeUpto array
SieveOfEratosthenes();
int low = 1, high = N, ans = 0;
while (low <= high) {
int mid = (low + high) >> 1;
// Check if the number is
// valid, try to reduce it
if (mid - primeUpto[mid] >= K) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
// ans is the minimum valid number
return (ans ? N - ans + 1 : 0);
}
// Driver Code
int main()
{
int N = 10, K = 3;
cout << countOfNumbers(N, K);
} Java
// Java implementation of the above approach
public class GFG{
static final int MAX = 1000001;
// primeUpto[i] denotes count of prime
// numbers upto i
static int primeUpto[] = new int [MAX];
// Function to compute all prime numbers
// and update primeUpto array
static void SieveOfEratosthenes()
{
int isPrime[] = new int[MAX];
for (int i=0; i < MAX ; i++ )
isPrime[i] = 1;
// 0 and 1 are not primes
isPrime[0] = isPrime[1] = 0;
for (int i = 2; i * i < MAX; i++) {
// If i is prime
if (isPrime[i] == 1) {
// Set all multiples of i as non-prime
for (int j = i * 2; j < MAX; j += i)
isPrime[j] = 0;
}
}
// Compute primeUpto array
for (int i = 1; i < MAX; i++) {
primeUpto[i] = primeUpto[i - 1];
if (isPrime[i] == 1)
primeUpto[i]++;
}
}
// Function to return the count
// of valid numbers
static int countOfNumbers(int N, int K)
{
// Compute primeUpto array
SieveOfEratosthenes();
int low = 1, high = N, ans = 0;
while (low <= high) {
int mid = (low + high) >> 1;
// Check if the number is
// valid, try to reduce it
if (mid - primeUpto[mid] >= K) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
ans = ans != 0 ? N - ans + 1 : 0 ;
// ans is the minimum valid number
return ans ;
}
// Driver Code
public static void main(String []args)
{
int N = 10, K = 3;
System.out.println(countOfNumbers(N, K)) ;
}
// This code is contributed by Ryuga
}Python3
# Python3 implementation of the above approach
MAX = 1000001
MAX_sqrt = MAX ** (0.5)
# primeUpto[i] denotes count of prime
# numbers upto i
primeUpto = [0] * (MAX)
# Function to compute all prime numbers
# and update primeUpto array
def SieveOfEratosthenes():
isPrime = [1] * (MAX)
# 0 and 1 are not primes
isPrime[0], isPrime[1] = 0, 0
for i in range(2, int(MAX_sqrt)):
# If i is prime
if isPrime[i] == 1:
# Set all multiples of i as non-prime
for j in range(i * 2, MAX, i):
isPrime[j] = 0
# Compute primeUpto array
for i in range(1, MAX):
primeUpto[i] = primeUpto[i - 1]
if isPrime[i] == 1:
primeUpto[i] += 1
# Function to return the count
# of valid numbers
def countOfNumbers(N, K):
# Compute primeUpto array
SieveOfEratosthenes()
low, high, ans = 1, N, 0
while low <= high:
mid = (low + high) >> 1
# Check if the number is
# valid, try to reduce it
if mid - primeUpto[mid] >= K:
ans = mid
high = mid - 1
else:
low = mid + 1
# ans is the minimum valid number
return (N - ans + 1) if ans else 0
# Driver Code
if __name__ == "__main__":
N, K = 10, 3
print(countOfNumbers(N, K))
# This code is contributed by Rituraj JainC#
// C# implementation of the above approach
using System;
public class GFG{
static int MAX = 1000001;
// primeUpto[i] denotes count of prime
// numbers upto i
static int []primeUpto = new int [MAX];
// Function to compute all prime numbers
// and update primeUpto array
static void SieveOfEratosthenes()
{
int []isPrime = new int[MAX];
for (int i=0; i < MAX ; i++ )
isPrime[i] = 1;
// 0 and 1 are not primes
isPrime[0] = isPrime[1] = 0;
for (int i = 2; i * i < MAX; i++) {
// If i is prime
if (isPrime[i] == 1) {
// Set all multiples of i as non-prime
for (int j = i * 2; j < MAX; j += i)
isPrime[j] = 0;
}
}
// Compute primeUpto array
for (int i = 1; i < MAX; i++) {
primeUpto[i] = primeUpto[i - 1];
if (isPrime[i] == 1)
primeUpto[i]++;
}
}
// Function to return the count
// of valid numbers
static int countOfNumbers(int N, int K)
{
// Compute primeUpto array
SieveOfEratosthenes();
int low = 1, high = N, ans = 0;
while (low <= high) {
int mid = (low + high) >> 1;
// Check if the number is
// valid, try to reduce it
if (mid - primeUpto[mid] >= K) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
ans = ans != 0 ? N - ans + 1 : 0 ;
// ans is the minimum valid number
return ans ;
}
// Driver Code
public static void Main()
{
int N = 10, K = 3;
Console.WriteLine(countOfNumbers(N, K)) ;
}
// This code is contributed by anuj_67..
}PHP
> 1;
// Check if the number is
// valid, try to reduce it
if ($mid - $primeUpto[$mid] >= $K)
{
$ans = $mid;
$high = $mid - 1;
}
else
$low = $mid + 1;
}
// ans is the minimum valid number
return ($ans ? $N - $ans + 1 : 0);
}
// Driver Code
$N = 10;
$K = 3;
echo countOfNumbers($N, $K);
// This code is contributed by mits
?>输出：
5