[Explanations] 8.10 Number of exams fight

Title Description

N numbers a1, a2, ..., an. EndSaH intend to choose the number 2 stitching.

A splice shows a x, y positive integers two strings as numbers, x y spliced into a new first numeric string after.

As will fight number 1234 with 56, to give a different result if obtained 123456. splicing sequence is not the same, as will the number of 56 and 1234 to fight, to give 561,234.

However, EndSaH not like a multiple of k, so he wants to know how many of a number of (i, j) (1 <= i, j <= n, i! = J) meet a [i] and a [j] after the number of fight, the result is not a multiple of k.

Input Format

The first line of two positive integers n, k, as defined problem.

The second behavior positive integer n, denotes the i-th a [i].

Output Format

Line a positive integer representing the number to meet the requirements of the number of

SAMPLE INPUT

4 11

45 1 10 12

Sample Output

data range

1<=n<=10^5,1<=k,a[i]<=10^9

Analytical thinking

We found that a [i] and the number of a [j] after splicing, is obtained:

a [i] * pow (10, len [j]) + a [j], len [j] represents a [j] the number of bits, such as a [j] = 45, len [j] = 2

If the result is a multiple of k, then there must be ((a [i] * pow (10, len [j]))% k + (a [j]% k))% k == 0.

Since a [i] up to 10, 10 so we can start the tub size k, coun [i] [j] indicates the length of i,% k = j is the number of how many, and then enumerate x, so t = a [i] * 10 ^ x% k. then there are coun [x] [kt] and the number is not a [i] stitching, in particular, for t = 0, there are coun [x] [0] a number and not a [i] stitching, so a [i] is the contribution of the answer n-num-1, num represents the number not a [i] stitching. Since k is too large, a map maintenance, O (n) enumerate a [i] to obtain the answer. Note 9 * 10 ^ 10 ^ 10 will burst long long, so to take the edge side mode.

Conferred on the Code

#include<bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;
const int N=1e5+5;
int n,k;
ll ans,a[N];
map<int,int> coun[15];
struct node{
  int len,mod;
}f[N];
inline int read(int now){
  int r=0,t=1,sum=0,c=getchar();
  while(c<'0'||c>'9') t=c=='-'?-1:1,c=getchar();
  while(c>='0'&&c<='9') r=r*10+c-48,sum++,c=getchar();
  f[now].len=sum;
  return r*t;
}
signed main(){
  //freopen("piece.in","r",stdin);
  //freopen("piece.out","w",stdout);
  int maxn=0;
  n=read(0),k=read(0);
  for(int i=1;i<=n;i++){
    a[i]=read(i);
    maxn=max(maxn,f[i].len);
  }
  for(int i=1;i<=n;i++){
    coun[f[i].len][a[i]%k]++;
    f[i].mod=a[i]%k;
  }
  ll sss;
  for(int i=1;i<=n;i++){
    int mt=a[i]%k;int num=0;
    coun[f[i].len][f[i].mod]--;
    for(int j=1;j<=maxn;j++){
      if(j>=7){
    sss=((ll)mt*(ll)pow(10,5))%k;
    sss=(sss*(ll)pow(10,j-5))%k;
      }
      else
    sss=((ll)mt*(ll)pow(10,j))%k;
      num+=coun[j][k-sss];
      if(sss==0) num+=coun[j][0];
    }
    ans+=n-num-1;
    coun[f[i].len][f[i].mod]++;
  }
  printf("%lld",ans);
  return 0;
}