CodeForces 55D Beautiful numbers（数位dp）

题意:

求[l,r]区间内 能被其各位数字(除0外)整除的 数的个数

分析:

dp[i][cur][lcm]:=i位,当前数,数字的最小公倍数,很显然的状态
但是cur很大1018,我们知道lcm{1⋯9}=2520,可以用这个压缩cur,存个mod
对于lcm我们发现能整除lcm的数字不多,打表发现也就50个,离散化一下
dp[i][mod][lcmid]:=这样这样状态20∗2520∗50∗(10决策数)就可以通过了

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PS:数位dp只用memset一次,状态可以重复利用−−之前T了一发

代码:

////  Created by TaoSama on 2015-10-19//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int __lcm(int a, int b) {    return a / __gcd(a, b) * b;}typedef long long LL;//lcm 1...9 = 2520const int MAGIC = 2520;LL dp[25][MAGIC][50], digit[25];int mp[MAGIC]; //total 48LL dfs(int i, int mod, int lcm, int e) {    if(!i) return mod % lcm == 0;    if(!e && ~dp[i][mod][mp[lcm]]) return dp[i][mod][mp[lcm]];    LL ret = 0;    int to = e ? digit[i] : 9;    for(int d = 0; d <= to; ++d) {        ret += dfs(i - 1, (mod * 10 + d) % MAGIC,                   d ? __lcm(lcm, d) : lcm, e && d == to);    }    if(!e) dp[i][mod][mp[lcm]] = ret;    return ret;}LL calc(LL x) {    int cnt = 0;    for(; x; x /= 10) digit[++cnt] = x % 10;    return dfs(cnt, 0, 1, 1);}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int cnt = 0;    for(int i = 1; i <= MAGIC; ++i)        if(MAGIC % i == 0) mp[i] = ++cnt;    memset(dp, -1, sizeof dp);    while(t--) {        LL l, r; scanf("%I64d%I64d", &l, &r);        printf("%I64d\n", calc(r) - calc(l - 1));    }    return 0;}