标签：node hang ica problem tin mes bit tip clu
http://codeforces.com/contest/1029/problem/D
You are given an array aa, consisting of nn positive integers.
Let‘s call a concatenation of numbers xx and yy the number that is obtained by writing down numbers xx and yy one right after another without changing the order. For example, a concatenation of numbers 1212 and 34563456 is a number 123456123456.
Count the number of ordered pairs of positions (i,j)(i,j) (i≠ji≠j) in array aa such that the concatenation of aiai and ajaj is divisible by kk.
The first line contains two integers nn and kk (1≤n≤2?1051≤n≤2?105, 2≤k≤1092≤k≤109).
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109).
Print a single integer — the number of ordered pairs of positions (i,j)(i,j) (i≠ji≠j) in array aa such that the concatenation of aiai and ajaj is divisible by kk.
6 11
45 1 10 12 11 7
7
4 2
2 78 4 10
12
5 2
3 7 19 3 3
0
代码：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int N, K;
int num[maxn];
map<long long, long long> mp[15];
int main() {
scanf("%d%d", &N, &K);
for(int i = 1; i <= N; i ++) {
scanf("%d", &num[i]);
long long a = num[i];
for(int j = 1; j <= 10; j ++) {
a *= 10;
a %= K;
mp[j][a] ++;
}
}
long long cnt = 0;
for(int i = 1; i <= N; i ++) {
int t = num[i] % K;
int len = log10(num[i]) + 1;
cnt += mp[len][(K - t) % K];
long long x = 1;
for(int j = 1; j <= len; j ++)
x = (x * 10) % K;
if(((num[i] * x) % K + num[i] % K) % K == 0)
cnt --;
}
printf("%I64d\n", cnt);
return 0;
}
CodeForces D. Concatenated Multiples
标签：node hang ica problem tin mes bit tip clu
原文地址：https://www.cnblogs.com/zlrrrr/p/9822824.html
