B. 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 ispalindrome 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,

Codeforces Round #449 (Div. 2) B. Chtholly's request【偶數位回文數】回文數系列題目(經典算法): http://blog.csdn.net/computer_liuyun/article/details/27967963

回文數系列題目(經典算法): http://blog.csdn.net/computer_liuyun/article/details/27967963

【題意】：求前k個偶數位回文數之和%p的值。

【分析】：POJ 1150/2402/3247同類題目。直接寫個判斷偶數位回文數的函數，将整數n逆轉，逆轉的方法就是利用求餘%和/的思想，此題總有一個常錯的地方，就是m的平方和立方可能會超範圍是以一定要用long long 或者_int64即可。或者用字元串表示，可能更加通俗易懂。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
ll n,m,sum;
ll a[15000];
ll huiwen(ll n)//倒過來和的數與原來相等，那麼就是回文數
{
    ll ans=n,t=n;//ll ans=n 若不分偶數位，則為ans=0;
    while(t)
    {
        ans=ans*10+t%10;
        t/=10;
    }
    return ans;
}

int main()
{
    cin>>n>>m;
    for(ll i=1;i<=n;i++)
    {
        sum=(sum+huiwen(i))%m;
    }
    printf("%lld\n",sum%m);
}

數字規律

#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
    cin.tie(0);\
    cout.tie(0);
using namespace std;
const int maxn = 1e5+10;
typedef long long LL;
typedef pair<int,int> P;
LL k,p;

LL getnow(LL n)
{
    char str[20] = {0};
    sprintf(str,"%lld",n);
    int len = strlen(str);
    for(int i=len;i<len+len;i++)
        str[i] = str[2*len-i-1];
    LL cnt = 0;
    sscanf(str,"%lld",&cnt);
    return cnt;
}

void solve()
{
    LL ans = 0;
    for(int i=1;i<=k;i++)
        ans = (ans+getnow(i)) % p;
//    cout<<getnow(k)<<endl;
    cout<<ans<<endl;
}

int main()
{
//    IO;
    cin>>k>>p;
    solve();
    return 0;
}

字元串

轉載于:https://www.cnblogs.com/Roni-i/p/7956422.html