给定数字N ，任务是找到所有N位数回文数的总和，这些数字可以被9整除，并且其中的数字不包含0。
注意：总和可能非常大，取10 9 +7为模。
例子：
Input: N = 2
Output: 99
Explanation:
There is only one 2-digit palindromic number divisible by 9 is 99.
Input: N = 3
Output: 4995
天真的方法：想法是遍历所有N位数字，并检查数字是否是回文数且可被9整除，并且其中是否不包含零。如果是，则所有数字的总和给出所需的结果。
高效方法：以下是对该问题陈述的一些观察：
- 如果N为奇数，则数字可以采用“ ab..x..ba”的形式；如果N为偶数，则数字可以为“ ab..xx..ba”的形式，其中’a’，’b’ x可以用1到9之间的数字代替。
- 如果数字“ ab..x..ba”可以被9整除，则(2 *(a + b)+ x)必须被9整除。
- 对于(a，b)的每个可能的对，总是存在一个这样的“ x”，因此该数字可以被9整除。
- 然后可以从下式计算可被9整除的N位回文数：
Since we have 9 digits to fill at the position a and b.
Therefore, total possible combinations will be:
if N is Odd then, count = pow(9, N/2)
if N is Even then, count = pow(9, N/2 - 1)
as N/2 digits are repeated at the end to get the
number palindromic.‘x’的唯一性证明：
- a和b的最小值为1，因此最小值之和= 2。
- a和b的最大值为9，因此最小值之和= 18。
- 由于数字是回文数，因此总和由下式给出：
sum = 2*(a+b) + x;- 2 *(a + b)的格式为2、4、6、8，…，36。为使和可被9整除，x的值可以是：
sum one of the possible value of x sum + x
2 7 9
4 5 9
6 3 9
8 1 9
. . .
. . .
. . .
36 9 45步骤如下：
- 经过上述观察，得出的所有N位回文数的位数在第k个位置可被9整除的总和为：
sum at kth position = 45*(number of combinations)/9- 循环遍历[1，N]并找到可能组合的数量，以在当前索引处放置一个数字。
- 使用上述公式找到当前位数的所有位数的总和。
- 在给定所需结果的情况下，每次迭代计算出的所有总和之和。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
long long int MOD = 1000000007;
// Function to find a^b efficiently
long long int power(long long int a,
long long int b)
{
// Base Case
if (b == 0)
return 1;
// To store the value of a^b
long long int temp = power(a, b / 2);
temp = (temp * temp) % MOD;
// Multiply temp by a until b is not 0
if (b % 2 != 0) {
temp = (temp * a) % MOD;
}
// Return the final ans a^b
return temp;
}
// Function to find sum of all N-digit
// palindromic number divisible by 9
void palindromicSum(int N)
{
long long int sum = 0, res, ways;
// Base Case
if (N == 1) {
cout << "9" << endl;
return;
}
if (N == 2) {
cout << "99" << endl;
return;
}
ways = N / 2;
// If N is even, decrease ways by 1
if (N % 2 == 0)
ways--;
// Find the total number of ways
res = power(9, ways - 1);
// Iterate over [1, N] and find the
// sum at each index
for (int i = 0; i < N; i++) {
sum = sum * 10 + 45 * res;
sum %= MOD;
}
// Print the final Sum
cout << sum << endl;
}
// Driver Code
int main()
{
int N = 3;
palindromicSum(N);
return 0;
} Java
// Java program for the above approach
class GFG{
static int MOD = 1000000007;
// Function to find a^b efficiently
static int power(int a, int b)
{
// Base Case
if (b == 0)
return 1;
// To store the value of a^b
int temp = power(a, b / 2);
temp = (temp * temp) % MOD;
// Multiply temp by a until b is not 0
if (b % 2 != 0)
{
temp = (temp * a) % MOD;
}
// Return the final ans a^b
return temp;
}
// Function to find sum of all N-digit
// palindromic number divisible by 9
static void palindromicSum(int N)
{
int sum = 0, res, ways;
// Base Case
if (N == 1)
{
System.out.print("9" + "\n");
return;
}
if (N == 2)
{
System.out.print("99" + "\n");
return;
}
ways = N / 2;
// If N is even, decrease ways by 1
if (N % 2 == 0)
ways--;
// Find the total number of ways
res = power(9, ways - 1);
// Iterate over [1, N] and find the
// sum at each index
for (int i = 0; i < N; i++)
{
sum = sum * 10 + 45 * res;
sum %= MOD;
}
// Print the final Sum
System.out.print(sum + "\n");
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
palindromicSum(N);
}
}
// This code is contributed by Amit KatiyarPython3
# Python 3 program for the above approach
MOD = 1000000007
# Function to find a^b efficiently
def power(a, b):
# Base Case
if (b == 0):
return 1
# To store the value of a^b
temp = power(a, b // 2)
temp = (temp * temp) % MOD
# Multiply temp by a until b is not 0
if (b % 2 != 0):
temp = (temp * a) % MOD
# Return the final ans a^b
return temp
# Function to find sum of all N-digit
# palindromic number divisible by 9
def palindromicSum(N):
sum = 0
# Base Case
if (N == 1):
print("9")
return
if (N == 2):
print("99")
return
ways = N // 2
# If N is even, decrease ways by 1
if (N % 2 == 0):
ways -= 1
# Find the total number of ways
res = power(9, ways - 1)
# Iterate over [1, N] and find the
# sum at each index
for i in range(N):
sum = sum * 10 + 45 * res
sum %= MOD
# Print the final Sum
print(sum)
# Driver Code
if __name__ == "__main__":
N = 3
palindromicSum(N)
# This code is contributed by chitranayalC#
// C# program for the above approach
using System;
class GFG{
static int MOD = 1000000007;
// Function to find a^b efficiently
static int power(int a, int b)
{
// Base Case
if (b == 0)
return 1;
// To store the value of a^b
int temp = power(a, b / 2);
temp = (temp * temp) % MOD;
// Multiply temp by a until b is not 0
if (b % 2 != 0)
{
temp = (temp * a) % MOD;
}
// Return the readonly ans a^b
return temp;
}
// Function to find sum of all N-digit
// palindromic number divisible by 9
static void palindromicSum(int N)
{
int sum = 0, res, ways;
// Base Case
if (N == 1)
{
Console.Write("9" + "\n");
return;
}
if (N == 2)
{
Console.Write("99" + "\n");
return;
}
ways = N / 2;
// If N is even, decrease ways by 1
if (N % 2 == 0)
ways--;
// Find the total number of ways
res = power(9, ways - 1);
// Iterate over [1, N] and find the
// sum at each index
for(int i = 0; i < N; i++)
{
sum = sum * 10 + 45 * res;
sum %= MOD;
}
// Print the readonly sum
Console.Write(sum + "\n");
}
// Driver Code
public static void Main(String[] args)
{
int N = 3;
palindromicSum(N);
}
}
// This code is contributed by AbhiThakur输出：
4995时间复杂度： O(N)，其中N是给定的数字。