chtholly’s-request

time limit per test 2 seconds

memory limit per test
256 megabytes
input
standard input
output
standard output

— Thanks a lot for today.

— I experienced so many great things.

— You gave me memories like dreams… But I have to leave now…

— One last request, can you…

— Help me solve a Codeforces problem?

— ……

— What?

Chtholly has been thinking about a problem for days:

If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.

Given integers k and p, calculate the sum of the k smallest zcy numbers and output this sum modulo p.

Unfortunately, Willem isn’t good at solving this kind of problems, so he asks you for help!
Input

The first line contains two integers k and p (1 ≤ k ≤ 105, 1 ≤ p ≤ 109).
Output

Output single integer — answer to the problem.
Examples
input

2 100

output

33

input

5 30

output

15

Note

In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.

In the second example, .

就是找位数是偶数的回文串，我一开始暴力判断每个数是不是回文，结果TLE

题解是 zcy=str+str的回文

从1-k枚举i，对于每个i，i接上i的回文就是一个zcy number，最大枚举到1e5即可

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#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
using namespace std;
const int maxn = 100010;
#define MAX 1e10
typedef long long LL;
long long zcy[100005];
LL (int a, int n);
int main() {
	
	LL k, p, i, j,num[maxn];
	cin >> k >> p;
	LL count = 0,ans=0;
	 for(i = 1; i < maxn-5; i++) {
		LL t = i, a = 0;
		int bit = 0;
		while (t) {
			a = a * 10 + t % 10;
			t /= 10;
			bit++;
		}
		t = i*pow(10, bit) + a;
		num[count++] = t;
	}
	 for (i = 0; i < k; i++)
		ans = (ans+num[i])%p;
	cout << ans << endl;
	
	return 0;
}
LL (int a, int n) {
	LL t = 1;
	while (n--) {
		t = t*a;
	}
	return t;
}