Codeforces 55D Beautiful Number
a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits.
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).
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).
Sample test(s)
Input
1
1 9
Output
9
Input
1
12 15
Output
2
思路:
- 在mod中，有一个规律，X%a = X%(b*a)%a; <=> X%( lcm(1,2,...,9) = 2520)%lcm(d[i]) == 0;即可将数值直接降到2520以内；
- 同时一个mod后的数，还需要记录的就是lcm(d[i]),这样每次又计算出来的子结构就直接相加即可(指mod之后的数值以及lcm相同的数(这时就可以看成是一个数))，lcm总共只有48个，(2^3,3^2,5,7的组合 4*3*2*2)；
- dp[i][j][k]: [i]: 高位为第i位，[j] : 在mod 2520之后的数值，[k]:记录下高位的lcm，由于直接来会MLE，所以离散化了(使用标号index[]);
代码思路参考了:AC_Von
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int (i)= 0;i < (n);i++)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
typedef long long ll;
const int MOD = ;
ll dp[][][];
int d[],index[MOD+];
void init()
{
for(int i = ,tot = ;i <= MOD;i++)
if(MOD % i == ) index[i] = tot++;
MS1(dp);
}
int lcm(int a,int b)
{
return a/__gcd(a,b)*b;
}
ll dfs(int pos,int prev,int prelcm,int edge)
{
if(pos == -) return prev % prelcm?:; // ***
ll ans = dp[pos][prev][index[prelcm]];
if( !edge && ~ans) return ans;
ans = ;
int e = edge ? d[pos]:;
for(int i = ;i <= e;i++){
int nowlcm = i ? lcm(prelcm,i) : prelcm;
int nowv = (prev * + i)%MOD;
ans += dfs(pos - ,nowv,nowlcm,edge && i == e);
}
if(!edge) dp[pos][prev][index[prelcm]] = ans;
return ans;
}
ll query(ll n)
{
MS0(d);int tot = ;
while(n){
d[tot++] = n%;
n /= ;
}
return dfs(tot - ,,,);
}
int main()
{
init();
int T;
cin>>T;
while(T--){
ll l,r;
scanf("%I64d%I64d",&l,&r);
printf("%I64d\n",query(r) - query(l-));
}
}