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Codeforces 55D Beautiful Number
a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits.
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 should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from li to ri, inclusively).
1
1 9
9
1
12 15
2
思路:
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 = 2520; ll dp[21][2520][50]; int d[21],index[MOD+5]; void init() { for(int i = 1,tot = 0;i <= MOD;i++) if(MOD % i == 0) 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 == -1) return prev % prelcm?0:1; // *** ll ans = dp[pos][prev][index[prelcm]]; if( !edge && ~ans) return ans; ans = 0; int e = edge ? d[pos]:9; for(int i = 0;i <= e;i++){ int nowlcm = i ? lcm(prelcm,i) : prelcm; int nowv = (prev * 10 + i)%MOD; ans += dfs(pos - 1,nowv,nowlcm,edge && i == e); } if(!edge) dp[pos][prev][index[prelcm]] = ans; return ans; } ll query(ll n) { MS0(d);int tot = 0; while(n){ d[tot++] = n%10; n /= 10; } return dfs(tot - 1,0,1,1); } int main() { init(); int T; cin>>T; while(T--){ ll l,r; scanf("%I64d%I64d",&l,&r); printf("%I64d\n",query(r) - query(l-1)); } }
Codeforces 55D Beautiful Number
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原文地址：http://www.cnblogs.com/hxer/p/5169877.html
