[CF55D]Beautiful numbers（数位dp，状态压缩）

题目链接：http://codeforces.com/problemset/problem/55/D

题意：给定区间，求区间内某数的所有数位能整除这个数本身的数的个数。

起初思路：dp(l,s,sum)表示这个数到l位，并且0~9出现的状态s，和为sum的时候的数字个数。发现这个sum不好处理，因为数字越来越大无法保证这个界限。用到一个性质：一个数能被一堆数整除，当且仅当这个数能被这堆数的最小公倍数整除。换句话说，我们统计这个数对1~9的最小公倍数取模，就能判断这个sum是否可以被各位数整除了。

结果MLE了，因为s这个状态统计了0~9 10个数的状态，所以占用空间是1024，因为0不用考虑，1一定能整除这个sum，所以删掉这两位。

#include <bits/stdc++.h>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const int maxn = 25;
const LL mod = 2520;
LL dp[maxn][1<<8][mod];
int digit[maxn];

LL lcm(LL x, LL y) {
  return x / __gcd(x, y) * y;
}

int get(int i, int s) {
  if(i < 2) return s;
  return s | (1 << (i-2));
}

LL dfs(int l, int s, int sum, bool flag) {
  if(l == 0) {
    For(i, 2, 10) {
      if(((1 << (i-2))& s) && (sum % i) != 0) return 0;
    }
    return 1;
  }
  if(!flag && ~dp[l][s][sum]) return dp[l][s][sum];
  LL ret = 0;
  int pos = !flag ? 9 : digit[l];
  Rep(i, pos+1) {
    ret += dfs(l-1, get(i, s), (sum*10+i)%mod, flag&&(i==pos));
  }
  if(!flag) dp[l][s][sum] = ret;
  return ret;
}

LL l, r;

LL f(LL x) {
  int pos = 0;
  while(x) {
    digit[++pos] = x % 10;
    x /= 10;
  }
  return dfs(pos, 0, 0, true);
}

signed main() {
  //FRead();
  Clr(dp, -1);
  int T;
  Rint(T);
  W(T) {
    cin >> l >> r;
    cout << f(r) - f(l-1) << endl;
  }
  RT 0;
}

One more time!

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int maxn = 22;
const int mod = 2520;
LL l, r;
LL dp[maxn][1<<8][mod];
int digit[maxn];

int get(int i, int st) {
    if(i < 2) return st;
    return st | (1 << (i - 2));
}

LL dfs(int l, int st, int sum, bool flag) {
    if(l == 0) {
        for(int i = 0; i <= 7; i++) {
            if(st & (1 << i) && sum % (i + 2) != 0) return 0; 
        }
        return 1;
    }
    if(!flag && ~dp[l][st][sum]) return dp[l][st][sum];
    int pos = flag ? digit[l] : 9;
    LL ret = 0;
    for(int i = 0; i <= pos; i++) {
        ret += dfs(l-1, get(i, st), (sum*10+i)%mod, flag&&(i==pos));
    }
    if(!flag) dp[l][st][sum] = ret;
    return ret;
}

LL f(LL x) {
    int pos = 0;
    while(x) {
        digit[++pos] = x % 10;
        x /= 10;
    }
    return dfs(pos, 0, 0, true);
}

int main() {
    // freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    memset(dp, -1, sizeof(dp));
    while(T--) {
        scanf("%lld%lld",&l,&r);
        printf("%lld
", f(r) - f(l-1));
    }
    return 0;
}