1-1 統計數字問題

問題描述

一本書的頁碼從自然數1 開始順序編碼直到自然數n。書的頁碼按照通常的習慣編排， 每個頁碼都不含多餘的前導數字0。例如，第6 頁用數字6 表示，而不是06 或006 等。數 字計數問題要求對給定書的總頁碼n，計算出書的全部頁碼中分別用到多少次數字0，1， 2，…，9。
給定表示書的總頁碼的10 進位制整數n (1≤n≤109${10}^9$) 。程式設計計算書的全部頁碼中分別用 到多少次數字0，1，2，…，9。

輸入

每個檔案只有1 行，給出表示書的總頁碼的整數n。

輸出

程式執行結束時，輸出檔案共有10行，在第k行 輸出頁碼中用到數字k-1 的次數，k=1，2，…，10。

Java: version-1

import java.io.*;

public class Main {

    public static void main(String[] args) {
        String input = "/Users/dijk/IdeaProjects/tongjishuziwenti/input.txt";
        String output = "/Users/dijk/IdeaProjects/tongjishuziwenti/output.txt";
        BufferedReader reader = null;
        BufferedWriter writer = null
 
;
        int n = 0;
        int i,j,v;
        int[] count = new int[10];

        try{
            //Read page number n from input.txt file
            reader = new BufferedReader(new FileReader(input));
            writer = new BufferedWriter(new FileWriter(output));
            String number = null;
            number = reader.readLine();
            if
 
(number != null){
                try{
                    n = Integer.parseInt(number);
                }catch (NumberFormatException e){
                    e.printStackTrace();
                }
            }

            //Count number 0-9 existed times
            for (i=1; i<=n; i++){
                j=i;
                do{
                    v = j%10;
                    j = j/10;
                    count[v]++;
                }while(j>0);
            }

            //Write number 0-9 and its existed times to output.txt
            for(int k=1; k<=10; k++){
                System.out.println(k-1 +":" + count[k-1]);
                writer.write(k-1 +":" + count[k-1] + "\n");
            }

            reader.close();
            writer.close();
        }catch(IOException e){
            e.printStackTrace();
        }finally {
            if(reader != null){
                try{
                    reader.close();
                }catch (IOException e){
                    e.printStackTrace();
                }
            }
            if(writer != null){
                try{
                    writer.close();
                }catch (IOException e){
                    e.printStackTrace();
                }
            }
        }
    }
}

Java: version-2

import java.io.*;

public class Main {

    public static void main(String[] args) {
        String input = "/Users/dijk/IdeaProjects/tongjishuziwenti/input.txt";
        String output = "/Users/dijk/IdeaProjects/tongjishuziwenti/output.txt";
        BufferedReader reader = null;
        BufferedWriter writer = null;
        int n = 0;
        int i,j;
        int[] count = new int[10];

        try{
            //Read page number n from input.txt file
            reader = new BufferedReader(new FileReader(input));
            writer = new BufferedWriter(new FileWriter(output));
            String number = null;
            number = reader.readLine();
            if(number != null){
                try{
                    n = Integer.parseInt(number);
                }catch (NumberFormatException e){
                    e.printStackTrace();
                }
            }

            //Count number 0-9 existed times
            int length = (int) Math.log10(n);//length的值為總頁碼數值n的位數減去1
            int curDigit, higher, lower;
            for(i=0; i<=length; i++){
                //eg: 如果 n=12345, curDigit=3，則higher=12，lower=45
                //則在數字3這個位置：
                //1、大於3的數字（4-9）出現的次數與前面的數字（higher）和該數字所在的位置（百位）有關：12*100=1200
                //2、數字3出現的次數與前面的數字（higher）、後面的數字（lower）和該數字所在的位置（百位）有關：12*100+45+1=1246
                //3、小於3的數字（0、1、2）出現的次數與前面的數字（higher）和該數字所在的位置（百位）有關：（12+1）*100=1300
                curDigit = (n/(int)Math.pow(10,length-i))%10;
                higher = n/(int)Math.pow(10,length-i+1);
                lower = n%(int)Math.pow(10,length-i);

                //equal current digit
                count[curDigit] += lower+1;

                //smaller than current digit
                for(j=0; j<curDigit; j++){
                    count[j] += (int)Math.pow(10,length-i)*(higher+1);
                }

                //equal and bigger than current digit
                for(j=curDigit; j<10; j++){
                    count[j] += (int)Math.pow(10,length-i)*higher;
                }
            }

            //減去多算的0
            for(i=0; i<=length; i++){
                count[0] -= (int)Math.pow(10,i);
            }

            //Write number 0-9 and its existed times to output.txt
            for(int k=1; k<=10; k++){
                System.out.println(k-1 +":" + count[k-1]);
                writer.write(k-1 +":" + count[k-1] + "\n");
            }

            reader.close();
            writer.close();
        }catch(IOException e){
            e.printStackTrace();
        }finally {
            if(reader != null){
                try{
                    reader.close();
                }catch (IOException e){
                    e.printStackTrace();
                }
            }
            if(writer != null){
                try{
                    writer.close();
                }catch (IOException e){
                    e.printStackTrace();
                }
            }
        }
    }
}

分析

由0，1，2，…，9組成的所有m位數。從m個0到m個9共有10m${10}^m$個m位數。在這10m${10}^m$個m位數中，包含數字的總個數是m∗10m$m\ast {10}^m$，其中0，1，2，…，9每個數字出現次數相同，則每個數字均為m∗10m−1$m\ast {10}^{m-1}$次。
可以首先把最高位的數字單獨處理，然後再處理剩下的m-1位，最後把那些多餘的0全部去掉。

Java: version-3

import java.io.*;

public class Main {

    public static void main(String[] args) {
        String input = "/Users/dijk/IdeaProjects/tongjishuziwenti/input.txt";
        String output = "/Users/dijk/IdeaProjects/tongjishuziwenti/output.txt";
        BufferedReader reader = null;
        BufferedWriter writer = null;
        int n = 0;
        int i,j,k;
        int[] count = new int[10];

        try{
            //Read page number n from input.txt file
            reader = new BufferedReader(new FileReader(input));
            writer = new BufferedWriter(new FileWriter(output));
            String number = null;
            number = reader.readLine();
            if(number != null){
                try{
                    n = Integer.parseInt(number);
                }catch (NumberFormatException e){
                    e.printStackTrace();
                }
            }

            //Count number 0-9 existed times
            int length = (int) Math.log10(n);//length的值為總頁碼數值n的位數減去1
            int tempNum = n;
            int highestDigit,restNum;
            for(i=0; i<=length; i++){
                highestDigit = tempNum/(int)Math.pow(10,length-i);//獲得最高位上的數字
                restNum = tempNum-highestDigit*(int)Math.pow(10,length-i);//去掉最高位剩下的數值
                count[highestDigit] += restNum+1;

                for(j=0; j<highestDigit; j++){
                    count[j] += (int)Math.pow(10,length-i);
                    for(k=0; k<10; k++){
                        count[k] += (length-i)*(int)Math.pow(10,length-i-1);
                    }
                }

                tempNum = restNum;
            }

            //減去多算的0
            for(i=0; i<=length; i++){
                count[0] -= (int)Math.pow(10,i);
            }

            //Write number 0-9 and its existed times to output.txt
            for(k=1; k<=10; k++){
                System.out.println(k-1 +":" + count[k-1]);
                writer.write(k-1 +":" + count[k-1] + "\n");
            }

            reader.close();
            writer.close();
        }catch(IOException e){
            e.printStackTrace();
        }finally {
            if(reader != null){
                try{
                    reader.close();
                }catch (IOException e){
                    e.printStackTrace();
                }
            }
            if(writer != null){
                try{
                    writer.close();
                }catch (IOException e){
                    e.printStackTrace();
                }
            }
        }
    }
}

Sample input 1

11

Sample output 1

0:1
1:4
2:1
3:1
4:1
5:1
6:1
7:1
8:1
9:1

Sample input 2

123

Sample output 2

0:22
1:57
2:27
3:23
4:22
5:22
6:22
7:22
8:22
9:22

Sample input 3

12345

Sample output 3

0:4664
1:8121
2:5121
3:4721
4:4671
5:4665
6:4664
7:4664
8:4664
9:4664

Reference

王曉東《計算機演算法設計與分析》（第3版）P6