HDU 6093 Rikka with Number （2017 Multi-Univ Training Contest 5）

Problem

d (d≥2)进制下的 好数 被定义为：K=(A1A2...Ad)d , 其中 Ai≠Aj 且 0≤Ai≤d−1 ，同时 A1≠0

求区间 [L,R] 中多少数恰好是 d 进制下的好数 ?

Limit

1≤L≤R≤105000

L,R 为十进制数

Idea

d 进制的最大好数值为 (d−1)⋅dd−1+(d−2)⋅dd−2+⋯+0⋅d0

d 进制的最小好数值为 1⋅dd−1+0⋅dd−2+2⋅dd−3+⋯+(d−1)⋅d0

故 d-1 进制下的最大好数与 d 进制下的最小好数满足：

(d−2)⋅(d−1)d−2+(d−3)⋅(d−1)d−3+⋯+0⋅(d−1)0<1⋅dd−1+0⋅dd−2+2⋅dd−3+⋯+(d−1)⋅d0

即 d 进制的好数区间与 d-1 进制的好数区间不会重叠。

故具体做法为：

STEP 1. 分别处理子问题 [1, L] 和 [1, R] 中有多少满足条件的好数

STEP 2. 二分枚举最大的 d 进制满足 d 的右区间仍 ≤N (N 作为区间的右值 [1, N]) 。

STEP 3. 在二分后，处理 d+1 进制中有多少满足条件的好数，额外添加即可。

STEP 4. 在处理 d+1 进制时，优先将 N 转换为 d+1 进制的数组，dfs 处理最大的 d+1 进制数。利用康托展开计算个数。

Code

import java.io.*;import java.math.BigInteger;import java.util.Scanner;public class HDU_6093 {    public static void main(String[] args) throws IOException {        SolverHDU_6093 solver = new SolverHDU_6093();        solver.run();    }}class SolverHDU_6093 {    private static final int MOD = 998244353;    private static final int N = 5000;    private int t;    private BigInteger L, R;    private long[] factorial = new long[N];    private int[] N2d = new int[N];    private boolean[] vis = new boolean[N];    private int[] maxDNum = new int[N];    private int[] bit = new int[N];    int lowbit(int x) {        return x & -x;    }    void add(int x) {        for (int i=x;i<N;i+=lowbit(i)) {            bit[i]++;        }    }    int get(int x) {        int res = 0;        for (int i=x;i!=0;i-=lowbit(i)) {            res += bit[i];        }        return res;    }    void init() {        factorial[0] = 1;        for (int i=1;i<N;i++) {            factorial[i] = factorial[i-1] * i % MOD;        }    }    boolean dfs(int idx, int d) {        if (idx > d)    return true;        if (vis[ N2d[idx] ] == true) {            for (int p=N2d[idx];p>=0;p--) {                if (vis[p] == false) {                    vis[p] = true;                    maxDNum[idx] = p;                    break;                } else if (p == 0) {                    return false;                }            }            for (int i=idx+1, p=d-1;i<=d;i++) {                while (vis[p])  p--;                maxDNum[i] = p;                vis[p] = true;            }            return true;        } else {            maxDNum[idx] = N2d[idx];            vis[ N2d[idx] ] = true;            if (dfs(idx+1, d) == false) {                vis[ N2d[idx] ] = false;                if (N2d[idx] == 0)  return false;                for (int p=N2d[idx]-1;p>=0;p--) {                    if (vis[p] == false) {                        vis[p] = true;                        maxDNum[idx] = p;                        break;                    } else if (p == 0) {                        return false;                    }                }                for (int i=idx+1, p=d-1;i<=d;i++) {                    while (vis[p])  p--;                    maxDNum[i] = p;                    vis[p] = true;                }                return true;            } else {                return true;            }        }    }    long calcExt(BigInteger N, int d) {        BigInteger NN = N;        for (int i=d;i!=0;i--) {            N2d[i] = (int) NN.mod(BigInteger.valueOf(d)).longValue();            NN = NN.divide(BigInteger.valueOf(d));        }        if (N2d[1] == 0)    return 0;        for (int i=0;i<=d;i++)  vis[i] = false;        vis[ N2d[1] ] = true;        maxDNum[1] = N2d[1];        if (dfs(2, d) == false) {            if (N2d[1] - 1 == 0)    return 0;            vis[ N2d[1] ] = false;            vis[ N2d[1] - 1 ] = true;            maxDNum[1] = N2d[1] - 1;            int p = d-1;            for (int i=2;i<=d;i++) {                while (vis[p] == true) p--;                maxDNum[i] = p;                vis[p] = true;            }        }        long ans = 0, dig;        for (int i=1;i<=d;i++)  bit[i] = 0;        for (int i=d;i!=0;i--) {            dig = get(maxDNum[i]);            add(maxDNum[i] + 1);            if (i == 1) dig--;            ans = (ans + dig * factorial[d-i]) % MOD;        }        return (ans+1) % MOD;    }    long calc(BigInteger N) {        int l = 2, r = 2000, mid, d = 1;        BigInteger maxNumInD;        while (l <= r) {            mid = (l+r) / 2;            maxNumInD = BigInteger.ZERO;            for (int j=mid-1;j>=0;j--) {                maxNumInD = maxNumInD.multiply(BigInteger.valueOf(mid));                maxNumInD = maxNumInD.add(BigInteger.valueOf(j));            }            if (maxNumInD.compareTo(N) <= 0) {                l = mid + 1;                d = mid;            } else {                r = mid - 1;            }        }        if (d == 1) return 0;        long ans = 0;        for (int i=2;i<=d;i++) {            ans += factorial[i] - factorial[i-1];            ans %= MOD;        }        ans += calcExt(N, d+1);        ans %= MOD;        return ans;    }    void run() throws IOException {        init();        Scanner cin = new Scanner(System.in);        t = cin.nextInt();        for (int ica=1;ica<=t;ica++) {            L = cin.nextBigInteger();            R = cin.nextBigInteger();            long left = calc(L.subtract(BigInteger.ONE));            long right = calc(R);            System.out.println((right - left + MOD) % MOD);        }    }}