LJX Campus: Entrance Ceremony (c + +)

LJX's campus: entrance ceremony

Difficulty level: C; operating time limit: 45ms; operating space limit: 256000KB; code length limit: 2000000B

Question Description

LJX in elementary School! He and YSM,YSF,WHT,LTJ and others are alumni. Today is a day of drowning in his life. Today, he will prove to his classmates that his maths is "happy". So, just learn simple a+b problem He, in class, to enemies Swatch Challenge Qaq, Swatch team have yzm,sjy,zzq and other people. And the LJX team has his good friends (JI) Friends: Ysm,ysf,wht,ltj,lzh and other people, the strength of the weak underestimated. There is one such problem: Given a positive integer n,m, the computational 1,2,......,n is required to be connected (1234567891011 ...). N) The value of mod m. "Recruit" Ljx thought, if M is 3, or 9, he will. But M can do anything, as long as it is less than INF. After the "appointment" of the various methods, the brain can't think out. So, right turned to (hehe) his (ZHU) teammates. But they had already run TAT. Jue Ruo LJX found you, he knelt and begged you to compile a program to help him solve the problem. Otherwise, he will "exercise" at the graduation ceremony and be qaq by the horrible swatch.

Input

* One line: positive integer n,m.

Output

* One line: output as required

Input example

Input Example 1 13 13 Input Example 2 12345678910 1000000000

Output example

Output Example 1 4 Output Example 2 345678910

Other Notes

N<=10^18 M<=10^9 This data is in the pit LJX AH Use today's matrix to learn the fast power Oh, no, it's a fast power for piecewise matrices.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

#define MAXN 4

using namespace Std;

typedef long long int LL;

int mod;

struct matrix
{
LL P[MAXN][MAXN];
}ans,tmp;

Matrix operator* (Matrix A,matrix B)
{
Matrix C;
for (int i=1;i<=3;i++)
for (int j=1;j<=3;j++)
{
c.p[i][j]=0;
for (int k=1;k<=3;k++)
C.p[i][j]= (c.p[i][j]+ (a.p[i][k]%mod) * (b.p[k][j]%mod)%mod)%mod;
}
return C;
}

void Cal (LL t,ll last)
{
memset (tmp.p,0,sizeof (TMP.P));
tmp.p[1][1]=t;
Tmp.p[2][1]=tmp.p[3][1]=tmp.p[2][2]=tmp.p[3][2]=tmp.p[3][3]=1;
LL y=last-t/10+1;
while (y)
{
if (y&1) ans=ans*tmp;
tmp=tmp*tmp;
y>>=1;
}
}

int main ()
{
for (int i=1;i<=3;i++)
Ans.p[i][i]=1;
LL N;
scanf ("%lld%lld", &n,&mod);
LL t=10;
while (n>=t)
{
Cal (T,t-1);
t*=10;
}
Cal (T,n);
printf ("%lld\n", ans.p[3][1]);
return 0;
}

LJX Campus: Entrance Ceremony (c + +)