HDU 6093

JAVA+大数搞了一遍- -

不是很麻烦- -

/*
HDU 6093 - Rikka with Number [ 进制转换,康托展开,大数 ]  |  2017 Multi-University Training Contest 5
题意：
	求L,R之间的好数的个数，好数要求在某个d(>=2)进制下数位是0到d-1的
分析：
	d 进制下好数的个数为 d!-(d-1)! ,且满足 d^(d-1) <= K <= d^d
	可知 若 N > d^d 则 1-N 包含前 d-1 个进制的所有好数
	寻找零界 d 可以二分，将 N 和 d^d 比大小
	
	计算临界的 d 进制的好数，用类似康托展开的过程
	即 对于 (x0, x1, x2...xn)，转化成 
	首位小于x0的所有排列+ 首位等于x0第二位小于x1的所有排列+....
	
*/
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
	static Scanner cin = new Scanner(System.in);
	static long MOD = 998244353;
	static BigInteger a[] = new BigInteger[2000];
	static long F[] = new long[2000];
	static BigInteger bb;
	static String s;
	static int [] b = new int[5005];
	static int [] c = new int[5005];
	static int lc, lb;
	static void init() {
		F[0] = 1;
		for (int i = 1; i < 2000; i++) F[i] = F[i-1] * i % MOD;
		for (int i = 1; i < 2000; i++) a[i] = BigInteger.valueOf(i).pow(i);	
	}
	static int BinaryFind(){
		int l = 2, r = 1999, mid = 0;
		while (l <= r) {
			mid = (l+r)/2;
			if (bb.compareTo(a[mid]) >= 0) l = mid+1;
			else r = mid-1;
		}
		return l;
	}
	static void change(int m) {
		lc = 0;
		while (lb > -0) {
			int x = 0;
			for (int i = lb-1; i >= 0; i--) {
				x = x*10 + b[i];
				b[i] = x/m;
				x %= m;
			}
			c[lc++] = x; while (lb > 0 && b[lb-1] == 0) lb--;
		}
	}
	static int[] vis = new int[5005];
	static long solve(){
		int m = BinaryFind();
		change(m);
		long ans = F[m-1] - F[1];
		if (m > lc) return ans;
		for (int i = 0; i < m; i++) vis[i] = 0;
		for (int i = 1; i < c[lc-1]; i++) 
			if (vis[i] == 0)
				ans = (ans + F[lc-1]) % MOD;
		vis[c[lc-1]] = 1;
		for (int i = lc-2; i >= 0; i--)
		{
			for (int j = 0; j < c[i]; j++)
				if (vis[j] == 0)
					ans = (ans + F[i]) % MOD;
			if (vis[c[i]] == 1) break;
			vis[c[i]] = 1;
			if (i == 0) ans = (ans + 1) % MOD;
		}
		return ans;
	}
	public static void main(String[] args) {
		init(); 
		int t = cin.nextInt();
		while ((t--) != 0) {		
			bb = cin.nextBigInteger();
			s = bb.toString();
			lb = s.length();
			for (int i = 0; i < lb; i++) {
				b[i] = (int)(s.charAt(lb-1-i) - '0');
			}
			b[0]--;
			for (int i = 0; i < lb; i++) {
				if (b[i] < 0) {
					b[i] += 10; b[i+1]--;
				}
				else break;
			}
			while (lb > 0 && b[lb-1] == 0) lb--;
			long ans = solve();
			bb = cin.nextBigInteger();
			s = bb.toString();
			lb = s.length();
			for (int i = 0; i < lb; i++) {
				b[i] = (int)(s.charAt(lb-1-i) - '0');
			}
			ans = (MOD + solve() - ans) % MOD;
			System.out.println(ans);
		}
	}
}