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2013
11-11

POJ 2282 The Counting Problem [解题报告] Java

The Counting Problem

问题描述 :

Given two integers a and b, we write the numbers between a and b, inclusive, in a list. Your task is to calculate the number of occurrences of each digit. For example, if a = 1024 and b = 1032, the list will be

1024 1025 1026 1027 1028 1029 1030 1031 1032


there are ten 0′s in the list, ten 1′s, seven 2′s, three 3′s, and etc.

输入:

The input consists of up to 500 lines. Each line contains two numbers a and b where 0 < a, b < 100000000. The input is terminated by a line `0 0', which is not considered as part of the input.

输出:

For each pair of input, output a line containing ten numbers separated by single spaces. The first number is the number of occurrences of the digit 0, the second is the number of occurrences of the digit 1, etc.

样例输入:

1 10
44 497
346 542
1199 1748
1496 1403
1004 503
1714 190
1317 854
1976 494
1001 1960
0 0

样例输出:

1 2 1 1 1 1 1 1 1 1
85 185 185 185 190 96 96 96 95 93
40 40 40 93 136 82 40 40 40 40
115 666 215 215 214 205 205 154 105 106
16 113 19 20 114 20 20 19 19 16
107 105 100 101 101 197 200 200 200 200
413 1133 503 503 503 502 502 417 402 412
196 512 186 104 87 93 97 97 142 196
398 1375 398 398 405 499 499 495 488 471
294 1256 296 296 296 296 287 286 286 247

解题代码:

/* @author: */
import java.util.*;
/**
 *
 * @author Leo
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void recurse(int [] digit,int n,int count){
        if(n<=0) return;
        int oneNo=n%10,tenNo;
        int m=n/10;
        tenNo=m;
        for(int i=0;i<=oneNo;i++){
            digit[i]+=count;
        }
        while(tenNo!=0){
            digit[tenNo%10]+=(oneNo+1)*count;
            tenNo/=10;
        }
        for(int i=0;i< 10;i++){
            digit[i]+=count*m;
        }
        digit[0]-=count;
        recurse(digit,m-1,10*count);
    }
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        while(true){
            int [] digita=new int[10];
            int [] digitb=new int[10];
            int a=sc.nextInt();
            int b=sc.nextInt();
            if(a==b&&b==0) break;
            if(a>b){
                int t=b;
                b=a;
                a=t;
            }
            recurse(digita,a-1,1);
            recurse(digitb,b,1);
            for(int i=0;i< 10;i++){
                System.out.print(Math.abs(digitb[i]-digita[i])+" ");
            }
            System.out.println();
        }
    }

}

  1. Gucci New Fall Arrivals

    This is really nice to know. I hope it will be successful in the future. Good job on this and keep up the good work.

  2. 思路二可以用一个长度为k的队列来实现,入队后判断下队尾元素的next指针是否为空,若为空,则出队指针即为所求。