一、题目描述

P8712 [蓝桥杯 2020 省 B1] 整数拼接

二、题目简析

我们选两个数 $a$ 和 $b$，用 $f(a,b)$ 表示 $a$ 在前、$b$ 在后的拼接，即 $f(a,b)=a\ast 10^{b.size}+b$。要满足 $k|f(a,b)$，则

$$\begin{array}{rl}f(a,b)& \equiv 0(\text{mod}k)\\ a\ast 10^{b.size}+b& \equiv 0(\text{mod}k)\\ a\ast 10^{b.size}& \equiv -b(\text{mod}k)\end{array}$$

三、AC代码

#include <bits/stdc++.h>

using namespace std;

const int MAX = 1e5 + 3;
typedef long long ll;
typedef pair<ll, int> P;

P A[MAX];
int n, k, R[13][MAX], fact[13];
ll ans; 

P quickin(void)
{
	ll ret = 0;
	int cnt = 0;
	bool flag = false;
	char ch = getchar();
	while (ch < '0' || ch > '9')
	{
		if (flag == '-')    flag = false;
		ch = getchar();
	}
	while ('0' <= ch && ch <= '9' && ch != EOF)
	{
		cnt++;
		ret = ret * 10 + ch - '0';
		ch = getchar();
	}
	if (flag)    ret = -ret;
	return P(ret, cnt);
}

ll quickme(ll x, ll n, ll m)
{
	ll ret = 0;
	while (n > 0)
	{
		if (n & 1)    ret = ret * x % m;
		x = x * x % m;
		n >>= 1;	
	}
	return ret;
}

int main()
{
	#ifdef LOCAL
	freopen("test.in", "r", stdin);
	#endif
	
	cin >> n >> k;
	for (int i = 0; i < n; i++)
	{
		A[i] = quickin();
		// 预处理 -b mod k 
		int res = (-A[i].first) % k;
		if (res < 0)    res += k;
		R[A[i].second][res]++;
	}
	
	fact[1] = 10 % k;
	for (int i = 2; i <= 10; i++)
		fact[i] = fact[i - 1] * 10 % k;
	
	for (int i = 0; i < n; i++)
	{
		int t = (-A[i].first) % k;
		if (t < 0)    t += k;
		for (int j = 1; j <= 10; j++)
		{
			// 求 a * 10^b.size mod k 
			int L = A[i].first * fact[j] % k;
			
			// 判重 
			if (A[i].second == j && t == L)
				ans += R[j][L] - 1;
			else
				ans += R[j][L];
		}
	}
	
	cout << ans << endl;
	
	return 0;	
}

完