D. Concatenated Multiples
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
In the first example pairs (1,2)(1,2), (1,3)(1,3), (2,3)(2,3), (3,1)(3,1), (3,4)(3,4), (4,2)(4,2), (4,3)(4,3) suffice. They produce numbers 451451, 45104510, 110110, 10451045, 10121012, 121121, 12101210, respectively, each of them is divisible by 1111.
In the second example all n(n−1)n(n−1) pairs suffice.
In the third example no pair is sufficient.
题意:给出n个数,再给一个mod,然后现在有一种方法说是,可以把任意两个数连接起来,问你连接起来的对数取余mod等于0的有多少个
思路:乍一看没什么思路,暴力肯定不行,10^5的数n^2就炸了,我们就要想怎么去优化他,我们可以考虑预处理然后遍历一遍
我们先给出一个例子 给出n个数和一个mod,求多少对加起来mod等于0?
这个n^2也不行,但是我们想一下如果我取余一个数=x 的话 我要什么情况才能让这个数加一个数%mod等于0
我们只有(x+y)%mod == 0 那在我们知道x的情况,我们只要找 mod-y的余数个数有多少即可
同理我们可以推理到这个题:因为是连接 12 34 我们相当于看成是 1200+34即可,就成功转移到了以上问题
因为连接不同的数的时候后面紧跟的0的个数不同,我们只要存下一个0到十个0所有的都用map存下即可
然后特别的,最后判断下自身加自身会不会也可以被mod为0,如果是的话-1
map<ll,ll> mp[11]
mp[j][x] 代表的是 j个0余数为x的个数
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