U Beautiful numbers 数位DP

题目

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

题解

%liuchanglc的博客

解题思路

这道题数据范围很大，暴力肯定会TLE
那我们就用一种高级的暴力——数位DP
顾名思义，就是在数位上做DP
如果不懂，在看看liuchanglc大佬的博客

代码

#include <cstdio>
#include <cstring>
#define int long long 
using namespace std;
int f[20][3000][60], a[60], m[3000], t;
int gcd(int x, int y) {
    return y ? gcd(y, x % y) : x;
}
int dp(int n, int s, int x, int g) {
    //g记录前一位能不能达到最大值
    if (n < 0) return (s % x) ? 0 : 1;
    if (!g && f[n][s][m[x]] != -1)
        return f[n][s][m[x]];
    int mn = 9, ms = 0;
    if (g) mn = a[n];
    for(int i = 0; i <= mn; i++)
        ms += dp(n - 1, (s *10 + i) % 2520, i ? i * x / gcd(i, x) : x, g && i == mn);
    if (!g) f[n][s][m[x]] = ms;
    return ms;
}
int getnum(int x) {
    int tot = 0;
    while (x) a[tot++] = x % 10, x /= 10;
    //a数组记录数字的每一位
    return dp(tot - 1, 0, 1, 1);
}
signed main() {
    scanf("%lld", &t);
    for(int i = 1, j = 0; i <= 2520; i++)
        if (2520 % i == 0) m[i] = j++;
    //m数组相当于离散化，压缩f数组的空间
    memset(f, -1, sizeof(f));
    while (t--) {
        int x, y;
        scanf("%lld%lld", &x, &y);
        printf("%lld
", getnum(y) - getnum(x - 1));
        //查分思想
    }
    return 0;
}