Codeforces55D Beautiful numbers

原题链接

虽然依旧是套模板，但是因为我太弱了，不会建状态，所以去看了题解。。

这里就直接引用我看的题解吧，写的不错的。

题解

//我的代码
#include<cstdio>
#include<cstring>
using namespace std;
const int mod = 2520;
const int N = mod + 10;
typedef long long ll;
int a[20], lc[N], l;
ll f[20][N][50];
inline ll re()
{
	ll x = 0;
	char c = getchar();
	bool p = 0;
	for (; c < '0' || c > '9'; c = getchar())
		p |= c == '-';
	for (; c >= '0' && c <= '9'; c = getchar())
		x = x * 10 + c - '0';
	return p ? -x : x;
}
int gcd(int x, int y)
{
	if (!y)
		return x;
	return gcd(y, x % y);
}
int LCM(int x, int y)
{
	if (!y)
		return x;
	return x / gcd(x, y) * y;
}
ll dfs(int pos, int nw, int lcm, int lm)
{
	if (pos < 0)
		return nw % lcm ? 0 : 1;
	if (!lm && f[pos][nw][lc[lcm]] > -1)
		return f[pos][nw][lc[lcm]];
	int i, k = lm ? a[pos] : 9;
	ll s = 0;
	for (i = 0; i <= k; i++)
		s += dfs(pos - 1, (nw * 10 + i) % mod, LCM(lcm, i), lm && i == a[pos]);
	if (!lm)
		return f[pos][nw][lc[lcm]] = s;
	return s;
}
ll calc(ll x)
{
	int k = 0;
	do
	{
		a[k++] = x % 10;
		x /= 10;
	} while (x > 0);
	return dfs(k - 1, 0, 1, 1);
}
int main()
{
	int i, t;
	ll n, m;
	for (i = 1; i * i <= mod; i++)
		if (!(mod % i))
		{
			lc[mod / i] = ++l;
			lc[i] = ++l;
		}
	memset(f, -1, sizeof(f));
	t = re();
	for (i = 1; i <= t; i++)
	{
		n = re();
		m = re();
		printf("%I64d\n", calc(m) - calc(n - 1));
	}
	return 0;
}