【题解】codeforces1029D[Codeforces Round #506 (Div. 3)]D.Concatenated Multiples 排序

Description

You are given an array $a$, consisting of $n$ positive integers.

Let’s call a concatenation of numbers $x$ and $y$ the number that is obtained by writing down numbers $x$ and $y$ one right after another without changing the order. For example, a concatenation of numbers $12$ and $3456$ is a number $123456$.

Count the number of ordered pairs of positions $(i,j) (i≠j)$ in array a such that the concatenation of $a_i$ and $a_j$ is divisible by $k$.

Input

The first line contains two integers $n$ and $k$ $(1≤n≤2\times10^5, 2≤k≤10^9)$.

The second line contains $n$ integers $a_1,a_2,\cdots,a_n$ $(1≤a_i≤10^9)$.

Output

Print a single integer — the number of ordered pairs of positions $(i,j) (i≠j)$ in array a such that the concatenation of $a_i$ and $a_j$ is divisible by $k$.

Examples

Input

6 11
45 1 10 12 11 7

Output

7

Input

4 2
2 78 4 10

Output

12

Input

5 2
3 7 19 3 3

Output

0

Note

In the first example pairs $(1,2), (1,3), (2,3), (3,1), (3,4), (4,2), (4,3)$ suffice. They produce numbers $451, 4510, 110, 1045, 1012, 121, 1210$, respectively, each of them is divisible by $11$.

In the second example all $n(n−1)$ pairs suffice.

In the third example no pair is sufficient.

---

记录每一个位数出现的模数，排序后统计每一个位数出现的需要的模数的个数，注意不要重复计算。

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=2e5+10;
int n,k,fac[11],len[N];
ll cnt,a[N];
vector<int>vx[11];
int main()
{
	//freopen("in.txt","r",stdin);
	scanf("%d%d",&n,&k);fac[0]=1;
	for(int i=1;i<=10;i++)fac[i]=fac[i-1]*10%k;
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		len[i]=log10(a[i])+1;
		vx[len[i]].push_back(a[i]%k);
	}
	for(int i=0;i<=10;i++)sort(vx[i].begin(),vx[i].end());
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=10;j++)
		{
			int mod=(a[i]*fac[j])%k;
			int key=(k-mod)%k;
			int l=lower_bound(vx[j].begin(),vx[j].end(),key)-vx[j].begin();
			int r=upper_bound(vx[j].begin(),vx[j].end(),key)-vx[j].begin();
			cnt+=r-l;
			if(len[i]==j&&(mod+a[i]%k)%k==0)cnt--;//去掉多算的 
		}
	}
	printf("%lld\n",cnt);
	return 0;
}

总结

这题看数据范围发现暴力不可解，可以在每一个位数下记录出现的 $k$ 模数，再进行计算。可以通过位数和模数来判断是否把自己给算进去了。