CodeForces 55D - Beautiful numbers

最新推荐文章于 2020-11-07 17:06:56 发布

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本文介绍了一种算法，用于计算给定范围内所有美丽数的数量。美丽数是指一个正整数，它能被其所有非零数字整除。文章提供了一个使用动态规划实现的C++程序示例。

Beautiful numbers

Time Limit:4000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

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 l_i and r_i (1 ≤ l_i ≤ r_i ≤ 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 l_i to r_i, inclusively).

Sample Input

Input

1
1 9

Output

Input

1
12 15

Output

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>

using namespace std;

typedef long long LL;
    
int a[20];
const int MOD = 2520;
LL dp[20][MOD][50];
int hash[MOD];

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 len, int mod, int lcm, bool limit)
{
    if(len < 1)    return mod%lcm==0;
    
    if(!limit && dp[len][mod][hash[lcm]] != -1)    return dp[len][mod][hash[lcm]];
    
    int maxn = limit ? a[len] : 9;
    LL ret = 0;
    int new_lcm;
    for(int i=0; i<=maxn; i++) {
        if(i)    new_lcm = LCM(lcm, i);
        else    new_lcm = lcm;
        ret += dfs(len-1, (mod*10+i)%MOD, new_lcm, limit&&i==maxn);
    }
    if(!limit)    dp[len][mod][hash[lcm]] = ret;
    return ret;
}

LL f(LL n)
{
    int len = 0;
    while(n) {
        a[++len] = n % 10;
        n/=10;
    }    
    return dfs(len, 0, 1, 1);
}

void Init()
{
    int cnt = 0;
    for(int i=1; i<=MOD; i++) {
        if(MOD % i == 0) {
            hash[i] = cnt++;
        }
    }
//    printf("cnt = %d\n", cnt); //
}

int main ()
{
    Init();
    int t;
    LL l, r;
    scanf("%d", &t);
    memset(dp, -1, sizeof(dp));
    while(t--) {
        scanf("%I64d%I64d", &l, &r);
        printf("%I64d\n", f(r)-f(l-1));        
    }

    return 0;
}