codeforce 897B_codeforces 897b-优快云博客

本文链接：https://blog.youkuaiyun.com/gtuif/article/details/78725662

本文探讨了如何解决一个特定的数学问题：寻找并计算前k个ZCY数的和，然后对该和进行模运算。ZCY数是指那些位数为偶数且正反读相同的数字。文章提供了一种直接模拟的方法来生成这些数，并给出了解决方案的代码实现。

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— 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

Input

5 30

Output

Note

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

In the second example, $\text{[math]}$ .

题意：如果一个数的位数是偶数并且正反读一样的话，那么就称这个数为zcy的数，给你k和p问你从最小的zcy数开始，连续k个数的和对p取余的结果；

思路：直接模拟就好了，首先要找到对应关系，1~11,2~22，。。。。。10~1001，所以第几小的数就是第几的回文串，所以相加取余即可；

下面附上我的代码：

/*1~11
  2~22
  3~33
  10~1001
  15~1551
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a[20];
LL k,p;
LL decimal(LL m)//求位数
{
    int res=0;
    while(m)
    {
        m/=10;
        res++;
    }
    return res;
}
LL solve(LL n)
{
    int p;
    int k=decimal(n);
    p=2*k-1;
    LL ans=0;
    while(n)
    {
        a[p-k]=a[k]=n%10;//直接赋成回文串
        n/=10;
        k++;
    }
   //for(int i=0;i<=p;i++)
     //   printf("%d",a[i]);
    //puts("");
    for(int i=0;i<=p;i++)
        ans=ans*10+a[i];
    //printf("%d ll%d\n",p,ans);
    return ans;
}
int main()
{

    while(cin>>k>>p)
    {
        LL s=0;
        for(int i=1;i<=k;i++)
                s=(s+solve(i))%p;//求和取余即可
        printf("%lld\n",s);
    }
    return 0;
}