Beautiful numbers
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).
Examples
Input
1
1 9
Output
9
Input
1
12 15
Output
2
一個數對它所有不爲0的位取模，結果都爲0，找出l到r區間所有這樣的數
- 數位dp，每一位求目前爲止這個數的大小，和模數，模數最大可能是2520
- 然後就得到dp[20][2521][2521], 就MLE了
- 雖然1-9的任意組合的最小公倍數最大爲2520，但是隻有48個模數
- 先求出所有的模數，然後再離散化，用mp[2521]映射到[49],這樣就不會MLE 了
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long num[100];
long long dp[100][2522][50];
long long ans[1200],mp[50000];
void init()
{
long long tail = -1;
for(long long i = 1; i<(1ll<<9); i++)
{
long long num = 1;
for(long long j = 0; j<9; j++)
{
if((i>>j)&1)
num = num*(j+1)/__gcd(num,j+1);
}
ans[++tail] = num;
}
sort(ans,ans+tail+1);
tail = unique(ans,ans+tail+1)-ans;
for(long long i = 0; i< tail; i++)
{
mp[ans[i]] = i;
// printf("%lld\n",ans[i]);
}
// printf("%lld\n",tail);
}
long long dfs(long long len,long long limit,long long sum,long long mod)
{
if(len==0)
return (sum%mod)==0?1:0;
if(!limit&&dp[len][sum][mp[mod]]!=-1)
return dp[len][sum][mp[mod]];
long long ans = 0;
long long up = limit?num[len]:9;
for(long long i = 0; i<=up; i++)
{
ans += dfs(len-1,limit&&i==up,(sum*10+i)%2520,i==0?mod:(mod==0?i:mod*i/__gcd(mod,i)));
}
if(!limit)
dp[len][sum][mp[mod]]=ans;
return ans;
}
long long solve(long long n)
{
long long tail = 0;
while(n)
{
num[++tail] = n%10;
n/=10;
}
return dfs(tail,1,0,1);
}
int main()
{
init();
long long n,m,t;
scanf("%lld",&t);
memset(dp,-1,sizeof(dp));
while(t--)
{
scanf("%lld %lld",&n,&m);
printf("%lld\n",solve(m)-solve(n-1));
}
return 0;
}