给定一个数组arr []和一个整数K ，任务是对索引对(i，j)进行计数，以使i！= j以及a [i]和a [j]的连接可被K整除。
例子：
Input: arr[] = [4, 5, 2], K = 2
Output: 4
Explanation:
All possible concatenations are {45, 42, 54, 52, 24, 25}.
Out of these, the numbers divisible by 2 are {42, 52, 24, 54}.
Therefore, the count is 4.
Input: arr[] =[45, 1, 10, 12, 11, 7], K = 11
Output: 7
天真的方法：
解决问题的最简单方法如下：
- 使用带有变量i和j的嵌套循环遍历数组。
- 对于每个i！= j ，通过等式连接arr [i]和arr [j] ：
Concatenation of arr[i] and arr[j] = (arr[i] * 10lenj) + arr[j], where lenj is the number of digits in arr[j]
- 用K检查连接数的可除性。
Concatenated Number is divisible by k if and only if the sum of (arr[j] % k) and ((arr[i] * 10lenj) % k) is 0 mod k.
- 对于所有这样的(arr [i]，arr [j])对，增加count 。
- 打印count的最终值。
时间复杂度： O(N 2 * len(maxm)，其中maxm表示数组中的最大元素，len(maxm)表示maxm的位数。
辅助空间： O(1)
高效方法：
要优化上述方法，请按照以下步骤操作：
- 要应用上述公式，请为1到10的每个长度维护一个Map。
- 在地图中存储{ len [a [i]] ， a [i]％k }。
- 计数对，对每个j在[1,10]，由(K – )％K((ARR [I] * 10 ^ j)的)的频率增加的计数存储在地图为{J，K – ((arr [i] * 10 ^ j)％k)}映射。
- 如果对(arr [i]，arr [i])进行计数，则将计数减少1 。
- 遍历数组后，打印最终计数。
下面是上述方法的实现：
C++
// C++ Program to count pairs
// of array elements which are
// divisible by K when concatenated
#include
using namespace std;
map rem[11];
// Function to calculate and return the
// count of pairs
int countPairs(vector a, int n, int k)
{
vector len(n);
// Compute power of 10 modulo k
vector p(11);
p[0] = 1;
for (int i = 1; i <= 10; i++) {
p[i] = (p[i - 1] * 10) % k;
}
for (int i = 0; i < n; i++) {
int x = a[i];
// Calculate length of a[i]
while (x > 0) {
len[i]++;
x /= 10;
}
// Increase count of remainder
rem[len[i]][a[i] % k]++;
}
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = 1; j <= 10; j++) {
// Calculate (a[i]* 10^lenj) % k
int r = (a[i] * p[j]) % k;
// Calculate (k - (a[i]* 10^lenj)% k) % k
int xr = (k - r) % k;
// Increase answer by count
ans += rem[j][xr];
// If a pair (a[i], a[i]) is counted
if (len[i] == j
&& (r + a[i] % k) % k == 0)
ans--;
}
}
// Return the count of pairs
return ans;
}
// Driver Code
int main()
{
vector a = { 4, 5, 2 };
int n = a.size(), k = 2;
cout << countPairs(a, n, k);
} Java
// Java program to count pairs
// of array elements which are
// divisible by K when concatenated
import java.util.*;
import java.lang.*;
class GFG{
static int[][] rem = new int[11][11];
// Function to calculate and return the
// count of pairs
static int countPairs(int[] a, int n, int k)
{
int[] len = new int[n];
// Compute power of 10 modulo k
int[] p = new int[11];
p[0] = 1;
for(int i = 1; i <= 10; i++)
{
p[i] = (p[i - 1] * 10) % k;
}
for(int i = 0; i < n; i++)
{
int x = a[i];
// Calculate length of a[i]
while (x > 0)
{
len[i]++;
x /= 10;
}
// Increase count of remainder
rem[len[i]][a[i] % k]++;
}
int ans = 0;
for(int i = 0; i < n; i++)
{
for(int j = 1; j <= 10; j++)
{
// Calculate (a[i]* 10^lenj) % k
int r = (a[i] * p[j]) % k;
// Calculate (k - (a[i]* 10^lenj)% k) % k
int xr = (k - r) % k;
// Increase answer by count
ans += rem[j][xr];
// If a pair (a[i], a[i]) is counted
if (len[i] == j &&
(r + a[i] % k) % k == 0)
ans--;
}
}
// Return the count of pairs
return ans;
}
// Driver code
public static void main (String[] args)
{
int[] a = { 4, 5, 2 };
int n = a.length, k = 2;
System.out.println(countPairs(a, n, k));
}
}
// This code is contributed by offbeatPython3
# Python3 program to count pairs
# of array elements which are
# divisible by K when concatenated
rem = [ [ 0 for x in range(11) ]
for y in range(11) ]
# Function to calculate and return the
# count of pairs
def countPairs(a, n, k):
l = [0] * n
# Compute power of 10 modulo k
p = [0] * (11)
p[0] = 1
for i in range(1, 11):
p[i] = (p[i - 1] * 10) % k
for i in range(n):
x = a[i]
# Calculate length of a[i]
while (x > 0):
l[i] += 1
x //= 10
# Increase count of remainder
rem[l[i]][a[i] % k] += 1
ans = 0
for i in range(n):
for j in range(1, 11):
# Calculate (a[i]* 10^lenj) % k
r = (a[i] * p[j]) % k
# Calculate (k - (a[i]* 10^lenj)% k) % k
xr = (k - r) % k
# Increase answer by count
ans += rem[j][xr]
# If a pair (a[i], a[i]) is counted
if (l[i] == j and
(r + a[i] % k) % k == 0):
ans -= 1
# Return the count of pairs
return ans
# Driver Code
a = [ 4, 5, 2 ]
n = len(a)
k = 2
print(countPairs(a, n, k))
# This code is contributed by chitranayalC#
// C# program to count pairs
// of array elements which are
// divisible by K when concatenated
using System;
class GFG{
static int [,]rem = new int[11, 11];
// Function to calculate and
// return the count of pairs
static int countPairs(int[] a,
int n, int k)
{
int[] len = new int[n];
// Compute power of 10 modulo k
int[] p = new int[11];
p[0] = 1;
for(int i = 1; i <= 10; i++)
{
p[i] = (p[i - 1] * 10) % k;
}
for(int i = 0; i < n; i++)
{
int x = a[i];
// Calculate length of a[i]
while (x > 0)
{
len[i]++;
x /= 10;
}
// Increase count of remainder
rem[len[i], a[i] % k]++;
}
int ans = 0;
for(int i = 0; i < n; i++)
{
for(int j = 1; j <= 10; j++)
{
// Calculate (a[i]* 10^lenj) % k
int r = (a[i] * p[j]) % k;
// Calculate (k - (a[i]* 10^lenj)% k) % k
int xr = (k - r) % k;
// Increase answer by count
ans += rem[j, xr];
// If a pair (a[i], a[i]) is counted
if (len[i] == j &&
(r + a[i] % k) % k == 0)
ans--;
}
}
// Return the count of pairs
return ans;
}
// Driver code
public static void Main(string[] args)
{
int[] a = {4, 5, 2};
int n = a.Length, k = 2;
Console.Write(countPairs(a, n, k));
}
}
// This code is contributed by rutvik_56输出
4
时间复杂度： O(N * len(maxm))
辅助空间： O(N)