The Counting Problem（区间数数，很好哦！）

超、凢脫俗 2022-08-02 13:53 208阅读 0赞

The Counting Problem

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;3000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;3680</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;1949</td> 
  </tr> 
 </tbody> 
</table>

Description

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.

Input

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.

Output

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.

Sample Input

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

Sample Output

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

Source

[Shanghai 2004][]

#include<cstdio>
    #include<iostream>
    #include<cstring>
    #define ll long long
    using namespace std;
    ll  b[12] = { 1, 10, 100, 1000, 10000, 100000, 1000000,10000000, 100000000, 1000000000, 10000000000, 100000000000 };
    ll count_num ( ll n, ll id )
    {
        ll left, m, sum = 0;
        for ( int i = 1; i < 12; i++ )
        {
            left = n / b[i] - (id==0);
            sum += left * b[i-1];
            m = (n % b[i] - n % b[i-1]) / b[i-1]; //求出从第到高的第i位上的具体数字
            if ( m > id ) sum += b[i-1];
            else if ( m == id ) sum += n % b[i-1] + 1;
            if ( n < b[i] ) break;
        }
        return sum;
    }
    int main()
    {
        ll n,m,f;
        ll a[10];
        int i;
        while(cin>>n>>m&&(m+n))
        {
            if(n>m)
            {
                f=n;
                n=m;
                m=f;
            }
            for(i=0;i<=9;i++)
            {
                a[i]=count_num(m,i)-count_num(n-1,i);
            }
            for(i=0;i<9;i++)
            {
                cout<<a[i]<<" ";
            }
            cout<<a[9]<<endl;
        }
    }

[Shanghai 2004]: http://poj.org/searchproblem?field=source&key=Shanghai+2004