Codeforces 55D Beautiful Numbers digital DP (Getting Started

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Test instructions

We think that a number num can be divisible by the number on each (expect 0) then this number is legal.

Give the interval [l,r] and ask how many valid numbers are in the interval.

First solve (long x) returns the number of legal numbers in [0, X], the answer is solve (r)-Solve (L-1);

Take 1234567 as an example

Flag indicates whether the current one can fill in, or can only fill in <= the corresponding number of digits

If the current search is third, and the second has been filled with 0 or 1, then the third place can be filled in at this time.

If the second fill is 2, then the third person can only fill in [0, 3], so DFS when a token, mark whether this can be filled.

If the current search is in the I-bit, it is clear that PRE_LCM and pre_mod are [highest bit, i+1] mod values and LCM

The function of Pre_mod is to determine whether the N-bit LCM is divisible by filling all the numbers (n-digits are filled).

If divisible, num = Lcm*x (and LCM is the largest LCM (1,2 9) = 2520, so 2520%LCM = = 0)

Set num = X*2520+y, then NUM%LCM = (x*2520+y)%LCM = Y%LCM

=>num%LCM = NUM%2520%LCM

And 2520 of all the factors may be 2~9 combination of common multiple, (of course, only part of the factor) Init () discretization of all the factors of 2520

DP[POS][PRE_MOD][PRE_LCM] represents the number of pre-POS digits to 2520 for Pre_mod and non-0-bit LCM bits PRE_LCM.

And the answer is the lowest back, so you can memorize it.

Import;import;import;import Java.math.biginteger;import jav<span style= "Background-color:rgb (255, 255, 255); > Xcode </span><span style= "font-family:arial, Helvetica, Sans-serif;" >a.text.DecimalFormat;</span>
Import Java.util.arraydeque;import java.util.arraylist;import Java.util.arrays;import Java.util.Collections;import Java.util.comparator;import Java.util.deque;import Java.util.hashmap;import Java.util.iterator;import Java.util.linkedlist;import;import Java.util.priorityqueue;import Java.util.Scanner;import Java.util.stack;import Java.util.stringtokenizer;import Java.util.treemap;import Java.util.TreeSet;import Java.util.queue;import;import;import;import;public class Main {int[] bit = new INT[30], hash = new int[2550];long[][][] dp = new long[30][255 0][50];long DFS (int pos, int pre_mod, int pre_lcm, Boolean flag) {if (pos = = 0) return pre_mod%pre_lcm==0?1:0;if (flag &AMP;&A mp DP[POS][PRE_MOD][HASH[PRE_LCM]]!=-1) return dp[pos][pre_mod][hash[pre_lcm]];int U = flag?9:bit[pos];long ans = 0L;for ( int d = 0; D <= u; d++) {int next_mod = (pre_mod*10+d)%mod;int NEXT_LCM = pre_lcm;if (d>0) NEXT_LCM = LCM (PRE_LCM, D); ans + = DFS (pos-1, Next_mod, NEXT_LCM, flag| | D&LT;U);} if (flag) DP[POS][PRE_MOD][HASH[PRE_LCM]] = Ans;return ans;} Long solve (long x) {int len = 0;while (x>0) {Bit[++len] = (int) (x%10l); x/=10l;} Return Dfs (len, 0, 1, false);} void Init () {int cnt = 0;for (int i = 1; I <= mod; i++) if (mod%i==0) hash[i]=++cnt;for (int i = 0; i <; i++) for (Int J = 0; J < 2520; J + +) for (int k = 0; k <; k++) dp[i][j][k] = 1;} void work () throws Exception{init (), int T = Int (), while (t-->0) {Long L = long (), R = Long (), Out.println (Solve (R)-Solve    (l-1l));}}    public static void Main (string[] args) throws exception{main wo = new main ();    in = new BufferedReader (new InputStreamReader (;  out = new PrintWriter (System.out);  in = new BufferedReader (new InputStreamReader (New FileInputStream (New File ("Input.txt")));        out = new PrintWriter (New File ("output.txt")); ();    Out.close (); }static int N = 3*100050;static int M = n*n * 10;D Ecimalformat df=new DecimalFormat ("0.0000"); static long inf = 1000000000000l;static long inf64 = (long) 1e18*2;static do  uble EPS = 1e-8;static double Pi = math.pi;static int mod = 2520;p rivate String Next () throws exception{while (str = = null | |    !str.hasmoreelements ()) str = new StringTokenizer (In.readline ());    return Str.nexttoken ();    } private int int () throws exception{return Integer.parseint (Next ());    } Private Long Long () throws exception{return Long.parselong (Next ());    } StringTokenizer str;    Static BufferedReader in;    static PrintWriter out; /*class Edge{int from, to, NEX; Edge () {}edge (int from, int. to, int nex) {This.from = From; = To;this.nex = NEX;}} Edge[] Edge = new edge[m<<1];int[] head = new Int[n];int edgenum;void Init_edge () {for (int i = 0; i < N; i++) head[ I] =-1; Edgenum = 0;} void Add (int u, int v) {Edge[edgenum] = new Edge (U, V, head[u]); Head[u] = edgenum++;} /**/int Upper_bound (int[] A, int l, int r, int val) {Upper_bound (a+l,a+r,val)-a;int pos = R;r--;while (l <= r) {int mid = (L + R) >> 1;if (A[mid] <= val) {L = m ID + 1;} else {pos = Mid;r = Mid-1;}} return POS;} int Pow (int x, int y) {int ans = 1;while (Y > 0) {if ((Y & 1) > 0) ans *= x;y >>= 1;x = x * x;} return ans;} Double Pow (double x, int y) {double ans = 1;while (Y > 0) {if ((Y & 1) > 0) ans *= x;y >>= 1;x = x * x;} return ans;} int pow_mod (int x, int y, int Mod) {int ans = 1;while (Y > 0) {if ((Y & 1) > 0) ans *= x;ans%= mod;y >>= 1 ; x = x * x;x%= mod;} return ans;} Long Pow (long x, long y) {long ans = 1;while (Y > 0) {if ((Y & 1) > 0) ans *= x;y >>= 1;x = x * x;} return ans;} Long Pow_mod (long x, long y, long Mod) {Long ans = 1;while (Y > 0) {if ((Y & 1) > 0) ans *= x;ans%= mod;y &GT;&G t;= 1;x = x * x;x%= mod;} return ans;} int Gcd (int x, int y) {if (x>y) {int tmp = x; x = y; y = tmp;} while (x>0) {y%= x;int tmp = x; x = y; y = tmp;} return y;} Long Gcd (long X, long y) {if (x>y) {long tmp = x; x = y; y = tmp;} while (x>0) {y%= x;long tmp = x; x = y; y = tmp;} return y;} int Lcm (int x, int y) {return x/gcd (x, y) *y;} Long Lcm (long x, long y) {return x/gcd (x, y) *y;} int max (int x, int y) {return x > y x:y;} int min (int x, int y) {return x < y x:y;} Double Max (double x, double y) {return x > y x:y;} Double min (double x, double y) {return x < y x:y;} Long Max (long x, long y) {return x > y x:y;} Long min (long x, long y) {return x < y x:y;} int abs (int x) {return x > 0 ×:-X;} Double abs (double x) {return x > 0 ×:-X;} Long ABS (long x) {return x > 0 ×:-X;} Boolean zero (double x) {return abs (x) < EPS;} Double sin (double x) {return Math.sin (x);} Double cos (double x) {return math.cos (x);} Double tan (double x) {return Math.tan (x);} Double sqrt (double x) {return math.sqrt (x);}}

Codeforces 55D Beautiful Numbers digital DP (Getting Started

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