CF55D：Beautiful numbers（数位dp + 数论）

reference：http://blog.youkuaiyun.com/lvshubao1314/article/details/43907225

D. Beautiful numbers

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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：求区间内有多少个数，其本身能被各个位上的数字整除。

思路：略。

# include <stdio.h>
# include <string.h>
# define LL long long
# define MOD 2520
LL a[21], index[2521], dp[21][MOD][50];
void init()
{
    int num = 0;
    for(int i=1; i<=MOD; ++i)
        if(!(MOD%i))
            index[i] = num++;//最小公倍数离散化，优化内存。
}

int gcd(int a, int b)
{
    if(b==0) return a;
    return gcd(b, a%b);
}

int lcm(int a, int b)
{
    return a/gcd(a, b)*b;
}

LL dfs(int pos, int sum, int prelcm, bool limit)
{
    if(pos == -1) return sum%prelcm==0;
    if(!limit && dp[pos][sum][index[prelcm]] != -1) return dp[pos][sum][index[prelcm]];
    int up = limit?a[pos]:9;
    LL ans = 0;
    for(int i=0; i<=up; ++i)
    {
        int newsum = (sum*10+i)%MOD;//2520为1~9的最小公倍数，数值对2520取余，如果该数（比2520大）能被其各位数整除，则它一定能被2520整除。
        int newlcm = prelcm;
        if(i)
            newlcm = lcm(newlcm, i);
        ans += dfs(pos-1, newsum, newlcm, limit&&(i==a[pos]));
    }
    if(!limit)
        dp[pos][sum][index[prelcm]] = ans;
    return ans;
}
LL solve(LL num)
{
    int len = 0;
    while(num)
    {
        a[len++] = num%10;
        num /= 10;
    }
    return dfs(len-1, 0, 1, true);
}
int main()
{
    LL n, m;
    int t;
    init();
    memset(dp, -1, sizeof(dp));
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d",&n,&m);
        printf("%I64d\n",solve(m)-solve(n-1));
    }
    return 0;
}