美团Codem编程比赛资格赛：数码

题目描述

CodeM美团点评编程大赛资格赛

时间限制：1秒

空间限制：32768K

给定两个整数 l 和 r ，对于所有满足1 ≤ l ≤ x ≤ r ≤ 10^9 的 x ，把 x 的所有约数全部写下来。对于每个写下来的数，只保留最高位的那个数码。求1～9每个数码出现的次数。

输入描述: 一行，两个整数 l 和 r (1 ≤ l ≤ r ≤ 10^9)。

输出描述: 输出9行。第 i 行，输出数码 i 出现的次数。

输入例子: 1 4

输出例子:

4
2
1
1
0
0
0
0
0

暴力法O(n)

改进的算法O(sqrt n)

import java.util.Scanner;

public class CodemDigits {
	
	public static void solve(int l, int r) {
		/* brute force
		long[] c1 = bruteForce(l-1, l-1);
		long[] c2 = bruteForce(r, r);
		for(int i=1; i<10; i++) {
			System.out.println(c2[i]-c1[i]);
		}
		*/
		long[] cl = countFactor(l-1);
		long[] cr = countFactor(r);
		for(int i=1; i<10; i++) {
			System.out.println(cr[i]-cl[i]);
		}
	}

	static int[] unit = {1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};

	public static long[] bruteForce(int upper, int maxFactor) {
		long[] c = new long[10];
		int unitIndex = 0;
		for(int factor=1; factor<=maxFactor; factor++) {
			int count = upper/factor;
			int highest = factor/unit[unitIndex];
			if(highest > 9) {
				unitIndex++;
				highest /= 10;
			}
			c[highest] += count;
		}
		return c;
	}

	public static long[] countHighest(int from, int to) {
		long[] c = new long[10];
		if(from > to) {
			return c;
		}
		if(to == 1000000000) {
			c[1]++;
			to -= 1;
		}
		for(int i=0; i to) break;
			if(from <= unit[i] && unit[i+1]-1 <= to) {
				for(int h=1; h<10; h++) {
					c[h] += unit[i];
				}
				continue;
			}
			for(int h=Math.max(from/unit[i], 1); h<10 && h<=to/unit[i]; h++) {
				c[h] += unit[i];
				if(h*unit[i] < from) {
					c[h] -= from - h*unit[i];
				}
				if((h+1)*unit[i]-1 > to) {
					c[h] -= (h+1)*unit[i] - 1 - to;
				}
			}
		}
		return c;
	}

	public static long[] countFactor(int upper) {
		if(upper <= 1000000) {
			return bruteForce(upper, upper);
		}
		
		long[] c = new long[10];
		
		/** count factor that appear < 1000 times **/
		//maxFactor[i] is the max factor that appears at least i times
		int[] maxFactor = new int[1001]; 
		for(int t=1; t<=1000; t++) {
			maxFactor[t] = upper/t;
		}
		
		for(int t=1; t<=999; t++) {
			long[] tmp = countHighest(maxFactor[t+1]+1, maxFactor[t]);
			for(int i=1; i<10; i++) {
				c[i] += tmp[i] * t;
			}
		}
		
		/** count factor that appear >= 1000 times **/
		long[] tmp = bruteForce(upper, maxFactor[1000]);

		for(int i=1; i<10; i++) {
			c[i] += tmp[i];
		}
		return c;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int l = sc.nextInt();
		int r = sc.nextInt();
		sc.close();
		solve(l,r);
	}

}