Codeforces Round #506 (Div. 3) 1029 D. Concatenated Multiples

题意：

　　给定n个数字，和一个模数k，从中选出两个数，直接拼接，问拼接成的数字是k的倍数的组合有多少个。

思路：

　　对于a，b两个数，假定$len = length of （b）$，那么a，b满足条件就是$a * （len个10） + b $是k的倍数，相当于$a * （len个10）% k + b % k  = k$；

那么我们可以预处理出每个数字%k的结果，用map计数。然后枚举每个数字，每个数字都有10种可能，因为len最大为10。每一次查找map中的数字要用find函数，不要用【】运算,find是二分，【】查找的复杂度高。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>


using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);



template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------showtime----------------------*/
        const int maxn = 2e5+9;
        ll a[maxn],lo[maxn];
        map<ll,ll>mp[11];
        ll n,k;
int main(){    
            scanf("%I64d%I64d", &n, &k);
            for(int i=1; i<=n; i++){
                
                scanf("%I64d", &a[i]);
                ll tmp = 1ll*a[i],len = 0;
                while(tmp > 0){
                    tmp/=10;
                    len++;
                }
                // debug(len);
                lo[i] = len;
                mp[len][a[i]%k]++;

            }
            ll ans = 0;
            for(int i=1; i<=n; i++){
                ll tmp = 1ll * a[i];
                // debug(tmp);
                for(int j=1; j<=10; j++){
                        tmp = (tmp * 10ll) % k;
                        ll c = (k-tmp);
                        if(c >= k) c = c % k;
                        //ans += mp[j][c];
                        auto po = mp[j].find(c);
                         if (po != mp[j].end()) ans += po->se;
                        if(lo[i] == j && a[i]%k == c)ans--;
                }
            }
            printf("%I64d\n",ans);

            return 0;
}