CodeForces 55D Beautiful numbers

最新推荐文章于 2024-06-17 14:54:52 发布

overload1997

最新推荐文章于 2024-06-17 14:54:52 发布

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分类专栏：数位dp

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本文介绍了一种使用数位动态规划的方法来解决一个特定问题：在给定区间内找出所有能够被其每一位非零数字整除的正整数的数量。通过状态压缩和数位DP技巧，文章详细阐述了解决方案的实现细节。

Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.

Input

The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two natural numbers li and ri (1 ≤ li ≤ ri ≤ 9 ·1018).

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

Output

Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from li to ri, inclusively).

   Examples
 
     input
   
1
1 9

     output
   
9

     input
   
1
12 15

     output
   
2

题意：求区间内能被自己各个位数整除的数有多少。

思路：一看就是数位dp，一个数能被各个位数整除也就是可以被最小公倍数整除，首先二进制状态压缩求出各2~9各个因数所组成的最小公倍数的所有情况，然后就是数位dp的思想，问题在于到了最后一位之后怎么确定这个数能被最小公倍数整除，我们记录两个数，一个是各位数的最小公倍数，另一个则是各位数%2520的情况，因为我们不可能记录整个数，所以我们就记录%2520之后的情况，剩下就没难度了。下面给代码：

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<bitset>
using namespace std;
#define maxn 2520
typedef long long LL;
LL dp[20][1 << 8][maxn];
int num[20], lcm[1 << 8];
int gcd(int a, int b){
	return !b ? a : gcd(b, a%b);
}
LL dfs(int pos, int status1,int status2, int limit){
	if (pos < 1){
		if (status1%lcm[status2])
			return 0;
		return 1;
	}
	if (!limit&&~dp[pos][status2][status1])
		return dp[pos][status2][status1];
	int end = limit ? num[pos] : 9;
	LL ans = 0;
	for (int i = 0; i <= end; i++){
		if (i < 2){
			ans += dfs(pos - 1, (status1 * 10 + i) % maxn, status2, limit&&i == end);
		}
		else
			ans += dfs(pos - 1, (status1 * 10 + i) % maxn, status2 | (1 << (i - 2)), limit&&i == end);
	}
	if (!limit){
		dp[pos][status2][status1] = ans;
	}
	return ans;
}
LL solve(LL x){
	int len = 0;
	while (x){
		num[++len] = x % 10;
		x /= 10;
	}
	return dfs(len, 0, 0, 1);
}
int main(){
	memset(dp, -1, sizeof(dp));
	for (int i = 0; i < (1 << 8); i++){
		lcm[i] = 1;
		for (int j = 0; (1 << j) <= i; j++){
			if ((1 << j)&i){
				int g = gcd(lcm[i], j + 2);
				lcm[i] = lcm[i] * (j + 2) / g;
			}
		}
	}
	int t;
	scanf("%d", &t);
	while (t--){
		LL l, r;
		scanf("%lld%lld", &l, &r);
		printf("%lld\n", solve(r) - solve(l - 1));
	}
}