给定一个数X ,任务是找到最小数N ,使得从1到N的所有因子的总和大于等于X。
例子:
Input: X = 10
Output: 5
Explanation:
Total factors of 1 = 1 (1)
Total factors of 2 = 2 (1, 2)
Total factors of 3 = 2 (1, 3)
Total factors of 4 = 3 (1, 2, 4)
Total factors of 5 = 2 (1, 5)
Total count = 1 + 2 + 2 + 3 + 2 = 10 which is is greater than or equal to X i.e., 10
Input: X = 19
Output: 8
Explanation:
Total factors of 1 = 1 (1)
Total factors of 2 = 2 (1, 2)
Total factors of 3 = 2 (1, 3)
Total factors of 4 = 3 (1, 2, 4)
Total factors of 5 = 2 (1, 5)
Total factors of 6 = 2 (1, 2, 3, 6)
Total factors of 7 = 2 (1, 7)
Total factors of 8 = 2 (1, 2, 4, 8)
Total count = 1 + 2 + 2 + 3 + 2 + 4 + 2 + 4 = 20 which is greater than or equal to X i.e., 19
天真的方法:天真的方法是从1开始运行一个循环,直到factor的总数大于X为止。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
#define MAX 1000050
#define lli long int
// Function to count total factors of N
lli CountFactors(lli N)
{
lli cnt = 0;
for (lli i = 1;
i * i <= N; i++) {
if (N % i == 0)
cnt += ((N / i == i) ? 1 : 2);
}
// Return the count
return cnt;
}
// Function to search lowest N
// such that the given condition
// is satisfied
lli minN(lli X)
{
lli i = 1;
lli total = 0;
while (1) {
// Add the count of total factor
// of i to variable total
total = total + CountFactors(i);
// If total count is greater than
// equal to N, then return i
if (total >= X)
return i;
i++;
}
}
// Driver Code
int main()
{
// Given sum
lli X = 10;
// Function Call
cout << minN(X) << endl;
return 0;
}
Java
// Java program for the above approach
class GFG{
static final int MAX = 1000050;
// Function to count total factors of N
static int CountFactors(int N)
{
int cnt = 0;
for(int i = 1; i * i <= N; i++)
{
if (N % i == 0)
cnt += ((N / i == i) ? 1 : 2);
}
// Return the count
return cnt;
}
// Function to search lowest N
// such that the given condition
// is satisfied
static int minN(int X)
{
int i = 1;
int total = 0;
while (true)
{
// Add the count of total factor
// of i to variable total
total = total + CountFactors(i);
// If total count is greater than
// equal to N, then return i
if (total >= X)
return i;
i++;
}
}
// Driver Code
public static void main(String[] args)
{
// Given sum
int X = 10;
// Function Call
System.out.print(minN(X) + "\n");
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 program for
# the above approach
MAX = 1000050
# Function to count
# total factors of N
def CountFactors(N):
cnt = 0
i = 1
while i * i <= N:
if (N % i == 0):
if (N // i == i):
cnt += 1
else:
cnt += 2
i += 1
# Return the count
return cnt
# Function to search lowest N
# such that the given condition
# is satisfied
def minN(X):
i = 1
total = 0
while (1):
# Add the count of total factor
# of i to variable total
total = total + CountFactors(i)
# If total count is greater than
# equal to N, then return i
if (total >= X):
return i
i += 1
# Driver Code
if __name__ == "__main__":
# Given sum
X = 10
# Function Call
print( minN(X))
# This code is contributed by Chitranayal
C#
// C# program for the above approach
using System;
class GFG{
//static readonly int MAX = 1000050;
// Function to count total factors of N
static int CountFactors(int N)
{
int cnt = 0;
for(int i = 1; i * i <= N; i++)
{
if (N % i == 0)
cnt += ((N / i == i) ? 1 : 2);
}
// Return the count
return cnt;
}
// Function to search lowest N
// such that the given condition
// is satisfied
static int minN(int X)
{
int i = 1;
int total = 0;
while (true)
{
// Add the count of total factor
// of i to variable total
total = total + CountFactors(i);
// If total count is greater than
// equal to N, then return i
if (total >= X)
return i;
i++;
}
}
// Driver Code
public static void Main(String[] args)
{
// Given sum
int X = 10;
// Function Call
Console.Write(minN(X) + "\n");
}
}
// This code is contributed by PrinciRaj1992
Javascript
C++
// C++ program for the above approach
#include
using namespace std;
#define MAX 1000050
#define lli long int
// Array to store smallest
// prime factors of each no.
lli spf[MAX + 1];
// Function to calculate smallest
// prime factor of N.
void calculate_SPF()
{
for (lli i = 0; i <= MAX; i++)
spf[i] = i;
for (lli i = 4; i <= MAX; i += 2)
spf[i] = 2;
for (lli i = 3;
i * i <= MAX; i++) {
if (spf[i] == i) {
for (int j = i * i;
j <= MAX; j += i)
// marking spf[j] if
// it is not previously
// marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Array to store the count of
// factor for N
lli tfactor[MAX + 1];
// Prefix array which contains
// the count of factors from 1 to N
lli pre[MAX + 1];
// Function to count total factors
// from 1 to N
void CountTotalfactors()
{
tfactor[1] = pre[1] = 1;
for (lli i = 2; i <= MAX; i++) {
lli mspf = spf[i];
lli prim = mspf;
lli temp = i;
lli cnt = 0;
while (temp % mspf == 0) {
temp /= mspf;
cnt += 1;
prim = prim * mspf;
}
// Store total factors of i
tfactor[i] = (cnt + 1)
* tfactor[temp];
// Stores total factors
// from 1 to i
pre[i] = pre[i - 1]
+ tfactor[i];
}
}
// Function to search lowest X
// such that the given condition
// is satisfied
lli BinarySearch(lli X)
{
lli start = 1;
lli end = MAX - 1;
while (start < end) {
// Find mid
lli mid = (start + end) / 2;
if (pre[mid] == X)
return mid;
// Search in the right half
else if (pre[mid] < X)
start = mid + 1;
// Search in the left half
else
end = mid;
}
// Return the position after
// Binary Search
return start;
}
// Function to find the required sum
void findSumOfCount(int X)
{
// Precompute smallest prime
// factor of each value
calculate_SPF();
// Calculate count of total
// factors from 1 to N
CountTotalfactors();
// Binary search to find minimum N
cout << BinarySearch(X)
<< endl;
}
// Driver Code
int main()
{
// Given Sum
int X = 10;
// Function Call
findSumOfCount(X);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
static final int MAX = 1000050;
// Array to store smallest
// prime factors of each no.
static int []spf = new int[MAX + 1];
// Function to calculate smallest
// prime factor of N.
static void calculate_SPF()
{
for(int i = 0; i <= MAX; i++)
spf[i] = i;
for(int i = 4; i <= MAX; i += 2)
spf[i] = 2;
for(int i = 3; i * i <= MAX; i++)
{
if (spf[i] == i)
{
for(int j = i * i;
j <= MAX; j += i)
// marking spf[j] if
// it is not previously
// marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Array to store the count of
// factor for N
static int []tfactor = new int[MAX + 1];
// Prefix array which contains
// the count of factors from 1 to N
static int []pre = new int[MAX + 1];
// Function to count total factors
// from 1 to N
static void CountTotalfactors()
{
tfactor[1] = pre[1] = 1;
for(int i = 2; i <= MAX; i++)
{
int mspf = spf[i];
int prim = mspf;
int temp = i;
int cnt = 0;
while (temp % mspf == 0)
{
temp /= mspf;
cnt += 1;
prim = prim * mspf;
}
// Store total factors of i
tfactor[i] = (cnt + 1) * tfactor[temp];
// Stores total factors
// from 1 to i
pre[i] = pre[i - 1] + tfactor[i];
}
}
// Function to search lowest X
// such that the given condition
// is satisfied
static int BinarySearch(int X)
{
int start = 1;
int end = MAX - 1;
while (start < end)
{
// Find mid
int mid = (start + end) / 2;
if (pre[mid] == X)
return mid;
// Search in the right half
else if (pre[mid] < X)
start = mid + 1;
// Search in the left half
else
end = mid;
}
// Return the position after
// Binary Search
return start;
}
// Function to find the required sum
static void findSumOfCount(int X)
{
// Precompute smallest prime
// factor of each value
calculate_SPF();
// Calculate count of total
// factors from 1 to N
CountTotalfactors();
// Binary search to find minimum N
System.out.print(BinarySearch(X) + "\n");
}
// Driver Code
public static void main(String[] args)
{
// Given Sum
int X = 10;
// Function Call
findSumOfCount(X);
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 program for the
# above approach
MAX = 1000050
# Array to store smallest
# prime factors of each no.
spf = [0 for i in range(MAX + 1)]
# Function to calculate smallest
# prime factor of N.
def calculate_SPF():
for i in range(MAX + 1):
spf[i] = i;
for i in range(4, MAX + 1, 2):
spf[i] = 2;
i = 3
while(i * i <= MAX):
if (spf[i] == i)
j = i * i
while(j <= MAX):
# Marking spf[j] if
# it is not previously
# marked
if (spf[j] == j):
spf[j] = i;
j += i
i += 1
# Array to store the
# count of factor for N
tfactor = [0 for i in range(MAX + 1)]
# Prefix array which contains
# the count of factors from 1 to N
pre = [0 for i in range(MAX + 1)]
# Function to count
# total factors from 1 to N
def CountTotalfactors():
tfactor[1] = pre[1] = 1;
for i in range(2, MAX + 1):
mspf = spf[i];
prim = mspf;
temp = i;
cnt = 0;
while (temp % mspf == 0):
temp //= mspf;
cnt += 1;
prim = prim * mspf;
# Store total factors of i
tfactor[i] = (cnt + 1) *
tfactor[temp];
# Stores total factors
# from 1 to i
pre[i] = pre[i - 1] +
tfactor[i];
# Function to search lowest X
# such that the given condition
# is satisfied
def BinarySearch(X):
start = 1;
end = MAX - 1;
while (start < end):
# Find mid
mid = (start + end) // 2;
if (pre[mid] == X):
return mid;
# Search in the right half
elif (pre[mid] < X):
start = mid + 1;
# Search in the left half
else:
end = mid;
# Return the position after
# Binary Search
return start;
# Function to find the
# required sum
def findSumOfCount(X):
# Precompute smallest prime
# factor of each value
calculate_SPF();
# Calculate count of total
# factors from 1 to N
CountTotalfactors();
# Binary search to find
# minimum N
print(BinarySearch(X))
# Driver code
if __name__ == "__main__":
# Given Sum
X = 10;
# Function Call
findSumOfCount(X);
# This code is contributed by rutvik_56
C#
// C# program for the above approach
using System;
class GFG{
const int MAX = 1000050;
// Array to store smallest
// prime factors of each no.
static int []spf = new int[MAX + 1];
// Function to calculate smallest
// prime factor of N.
static void calculate_SPF()
{
for(int i = 0; i <= MAX; i++)
spf[i] = i;
for(int i = 4; i <= MAX; i += 2)
spf[i] = 2;
for(int i = 3; i * i <= MAX; i++)
{
if (spf[i] == i)
{
for(int j = i * i;
j <= MAX; j += i)
// marking spf[j] if
// it is not previously
// marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Array to store the count of
// factor for N
static int []tfactor = new int[MAX + 1];
// Prefix array which contains
// the count of factors from 1 to N
static int []pre = new int[MAX + 1];
// Function to count total factors
// from 1 to N
static void CountTotalfactors()
{
tfactor[1] = pre[1] = 1;
for(int i = 2; i <= MAX; i++)
{
int mspf = spf[i];
int prim = mspf;
int temp = i;
int cnt = 0;
while (temp % mspf == 0)
{
temp /= mspf;
cnt += 1;
prim = prim * mspf;
}
// Store total factors of i
tfactor[i] = (cnt + 1) * tfactor[temp];
// Stores total factors
// from 1 to i
pre[i] = pre[i - 1] + tfactor[i];
}
}
// Function to search lowest X
// such that the given condition
// is satisfied
static int BinarySearch(int X)
{
int start = 1;
int end = MAX - 1;
while (start < end)
{
// Find mid
int mid = (start + end) / 2;
if (pre[mid] == X)
return mid;
// Search in the right half
else if (pre[mid] < X)
start = mid + 1;
// Search in the left half
else
end = mid;
}
// Return the position after
// Binary Search
return start;
}
// Function to find the required sum
static void findSumOfCount(int X)
{
// Precompute smallest prime
// factor of each value
calculate_SPF();
// Calculate count of total
// factors from 1 to N
CountTotalfactors();
// Binary search to find minimum N
Console.Write(BinarySearch(X) + "\n");
}
// Driver Code
public static void Main()
{
// Given Sum
int X = 10;
// Function Call
findSumOfCount(X);
}
}
// This code is contributed by Code_Mech
5
时间复杂度: O(N * sqrt(N))
辅助空间: O(1)
高效方法:可以通过进行一些预计算来优化上述方法。步骤如下:
- 使用本文讨论的方法计算并存储每个数字的最小素因数(SPF)。
- 使用Eratosthenes筛和上述计算的最小素因数,可以有效地找到N的因数。
- 创建一个前缀和数组,以存储从1到MAX的因子计数之和。
- 为了找到满足上述条件的最低N ,而不是在前缀数组中进行线性搜索,而是对前缀和数组进行二进制搜索(因为前缀和数组的顺序将递增),因此时间复杂度对于多个查询也。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
#define MAX 1000050
#define lli long int
// Array to store smallest
// prime factors of each no.
lli spf[MAX + 1];
// Function to calculate smallest
// prime factor of N.
void calculate_SPF()
{
for (lli i = 0; i <= MAX; i++)
spf[i] = i;
for (lli i = 4; i <= MAX; i += 2)
spf[i] = 2;
for (lli i = 3;
i * i <= MAX; i++) {
if (spf[i] == i) {
for (int j = i * i;
j <= MAX; j += i)
// marking spf[j] if
// it is not previously
// marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Array to store the count of
// factor for N
lli tfactor[MAX + 1];
// Prefix array which contains
// the count of factors from 1 to N
lli pre[MAX + 1];
// Function to count total factors
// from 1 to N
void CountTotalfactors()
{
tfactor[1] = pre[1] = 1;
for (lli i = 2; i <= MAX; i++) {
lli mspf = spf[i];
lli prim = mspf;
lli temp = i;
lli cnt = 0;
while (temp % mspf == 0) {
temp /= mspf;
cnt += 1;
prim = prim * mspf;
}
// Store total factors of i
tfactor[i] = (cnt + 1)
* tfactor[temp];
// Stores total factors
// from 1 to i
pre[i] = pre[i - 1]
+ tfactor[i];
}
}
// Function to search lowest X
// such that the given condition
// is satisfied
lli BinarySearch(lli X)
{
lli start = 1;
lli end = MAX - 1;
while (start < end) {
// Find mid
lli mid = (start + end) / 2;
if (pre[mid] == X)
return mid;
// Search in the right half
else if (pre[mid] < X)
start = mid + 1;
// Search in the left half
else
end = mid;
}
// Return the position after
// Binary Search
return start;
}
// Function to find the required sum
void findSumOfCount(int X)
{
// Precompute smallest prime
// factor of each value
calculate_SPF();
// Calculate count of total
// factors from 1 to N
CountTotalfactors();
// Binary search to find minimum N
cout << BinarySearch(X)
<< endl;
}
// Driver Code
int main()
{
// Given Sum
int X = 10;
// Function Call
findSumOfCount(X);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
static final int MAX = 1000050;
// Array to store smallest
// prime factors of each no.
static int []spf = new int[MAX + 1];
// Function to calculate smallest
// prime factor of N.
static void calculate_SPF()
{
for(int i = 0; i <= MAX; i++)
spf[i] = i;
for(int i = 4; i <= MAX; i += 2)
spf[i] = 2;
for(int i = 3; i * i <= MAX; i++)
{
if (spf[i] == i)
{
for(int j = i * i;
j <= MAX; j += i)
// marking spf[j] if
// it is not previously
// marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Array to store the count of
// factor for N
static int []tfactor = new int[MAX + 1];
// Prefix array which contains
// the count of factors from 1 to N
static int []pre = new int[MAX + 1];
// Function to count total factors
// from 1 to N
static void CountTotalfactors()
{
tfactor[1] = pre[1] = 1;
for(int i = 2; i <= MAX; i++)
{
int mspf = spf[i];
int prim = mspf;
int temp = i;
int cnt = 0;
while (temp % mspf == 0)
{
temp /= mspf;
cnt += 1;
prim = prim * mspf;
}
// Store total factors of i
tfactor[i] = (cnt + 1) * tfactor[temp];
// Stores total factors
// from 1 to i
pre[i] = pre[i - 1] + tfactor[i];
}
}
// Function to search lowest X
// such that the given condition
// is satisfied
static int BinarySearch(int X)
{
int start = 1;
int end = MAX - 1;
while (start < end)
{
// Find mid
int mid = (start + end) / 2;
if (pre[mid] == X)
return mid;
// Search in the right half
else if (pre[mid] < X)
start = mid + 1;
// Search in the left half
else
end = mid;
}
// Return the position after
// Binary Search
return start;
}
// Function to find the required sum
static void findSumOfCount(int X)
{
// Precompute smallest prime
// factor of each value
calculate_SPF();
// Calculate count of total
// factors from 1 to N
CountTotalfactors();
// Binary search to find minimum N
System.out.print(BinarySearch(X) + "\n");
}
// Driver Code
public static void main(String[] args)
{
// Given Sum
int X = 10;
// Function Call
findSumOfCount(X);
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 program for the
# above approach
MAX = 1000050
# Array to store smallest
# prime factors of each no.
spf = [0 for i in range(MAX + 1)]
# Function to calculate smallest
# prime factor of N.
def calculate_SPF():
for i in range(MAX + 1):
spf[i] = i;
for i in range(4, MAX + 1, 2):
spf[i] = 2;
i = 3
while(i * i <= MAX):
if (spf[i] == i)
j = i * i
while(j <= MAX):
# Marking spf[j] if
# it is not previously
# marked
if (spf[j] == j):
spf[j] = i;
j += i
i += 1
# Array to store the
# count of factor for N
tfactor = [0 for i in range(MAX + 1)]
# Prefix array which contains
# the count of factors from 1 to N
pre = [0 for i in range(MAX + 1)]
# Function to count
# total factors from 1 to N
def CountTotalfactors():
tfactor[1] = pre[1] = 1;
for i in range(2, MAX + 1):
mspf = spf[i];
prim = mspf;
temp = i;
cnt = 0;
while (temp % mspf == 0):
temp //= mspf;
cnt += 1;
prim = prim * mspf;
# Store total factors of i
tfactor[i] = (cnt + 1) *
tfactor[temp];
# Stores total factors
# from 1 to i
pre[i] = pre[i - 1] +
tfactor[i];
# Function to search lowest X
# such that the given condition
# is satisfied
def BinarySearch(X):
start = 1;
end = MAX - 1;
while (start < end):
# Find mid
mid = (start + end) // 2;
if (pre[mid] == X):
return mid;
# Search in the right half
elif (pre[mid] < X):
start = mid + 1;
# Search in the left half
else:
end = mid;
# Return the position after
# Binary Search
return start;
# Function to find the
# required sum
def findSumOfCount(X):
# Precompute smallest prime
# factor of each value
calculate_SPF();
# Calculate count of total
# factors from 1 to N
CountTotalfactors();
# Binary search to find
# minimum N
print(BinarySearch(X))
# Driver code
if __name__ == "__main__":
# Given Sum
X = 10;
# Function Call
findSumOfCount(X);
# This code is contributed by rutvik_56
C#
// C# program for the above approach
using System;
class GFG{
const int MAX = 1000050;
// Array to store smallest
// prime factors of each no.
static int []spf = new int[MAX + 1];
// Function to calculate smallest
// prime factor of N.
static void calculate_SPF()
{
for(int i = 0; i <= MAX; i++)
spf[i] = i;
for(int i = 4; i <= MAX; i += 2)
spf[i] = 2;
for(int i = 3; i * i <= MAX; i++)
{
if (spf[i] == i)
{
for(int j = i * i;
j <= MAX; j += i)
// marking spf[j] if
// it is not previously
// marked
if (spf[j] == j)
spf[j] = i;
}
}
}
// Array to store the count of
// factor for N
static int []tfactor = new int[MAX + 1];
// Prefix array which contains
// the count of factors from 1 to N
static int []pre = new int[MAX + 1];
// Function to count total factors
// from 1 to N
static void CountTotalfactors()
{
tfactor[1] = pre[1] = 1;
for(int i = 2; i <= MAX; i++)
{
int mspf = spf[i];
int prim = mspf;
int temp = i;
int cnt = 0;
while (temp % mspf == 0)
{
temp /= mspf;
cnt += 1;
prim = prim * mspf;
}
// Store total factors of i
tfactor[i] = (cnt + 1) * tfactor[temp];
// Stores total factors
// from 1 to i
pre[i] = pre[i - 1] + tfactor[i];
}
}
// Function to search lowest X
// such that the given condition
// is satisfied
static int BinarySearch(int X)
{
int start = 1;
int end = MAX - 1;
while (start < end)
{
// Find mid
int mid = (start + end) / 2;
if (pre[mid] == X)
return mid;
// Search in the right half
else if (pre[mid] < X)
start = mid + 1;
// Search in the left half
else
end = mid;
}
// Return the position after
// Binary Search
return start;
}
// Function to find the required sum
static void findSumOfCount(int X)
{
// Precompute smallest prime
// factor of each value
calculate_SPF();
// Calculate count of total
// factors from 1 to N
CountTotalfactors();
// Binary search to find minimum N
Console.Write(BinarySearch(X) + "\n");
}
// Driver Code
public static void Main()
{
// Given Sum
int X = 10;
// Function Call
findSumOfCount(X);
}
}
// This code is contributed by Code_Mech
5
时间复杂度: O(MAX * log(MAX))
辅助空间: O(MAX)
注意:如果存在Q个查询,则为所有Q个查询找到N的时间复杂度将为O(Q * log(MAX)) 。