UVA - 1225 Digit Counting

Description

Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequence of consecutive integers starting with 1 to N (1 < N < 10000). After that, he counts the number of times each digit (0 to 9) appears in the sequence. For example, with N = 13 , the sequence is:

12345678910111213

In this sequence, 0 appears once, 1 appears 6 times, 2 appears 2 times, 3 appears 3 times, and each digit from 4 to 9 appears once. After playing for a while, Trung gets bored again. He now wants to write a program to do this for him. Your task is to help him with writing this program.

Input 

The input file consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets.

For each test case, there is one single line containing the number N .

Output 

For each test case, write sequentially in one line the number of digit 0, 1,...9 separated by a space.

Sample Input 

2 
3 
13

Sample Output 

0 1 1 1 0 0 0 0 0 0 
1 6 2 2 1 1 1 1 1 1

把前n个整数依次写在一起，数一数0~9各出现多少次。输出十个整数。以空格间隔。 n小于一万。

先打表记录每个n对应的结果，然后直接输出结果。

[cpp]  view plain  copy

1. #include <iostream>  
2. #include<cstring>  
3. #define maxn 10005  
4. using namespace std;  
5. int main(){  
6. int T;  
7. cin>>T;  
8. int n;  
9. int ans[maxn][10];  
10. memset(ans,0,sizeof(ans));  
11.   
12. for(int i=1;i<maxn;i++)  
13.     {  
14.        int m=i;  
15.        if(i!=1) {for(int j=0;j<10;j++)//后一个数字只需要在前一个数字的基础上增加当前数字对数组所作出的贡献即可  
16.         ans[i][j]=ans[i-1][j];}  
17.     while(m>0)//分别记录每位的数字  
18.     {  
19.     ans[i][m%10]++; m/=10;  
20.     }  
21.   
22.     }  
23. while(T--)  
24. {  
25.     cin>>n;  
26.     cout<<ans[n][0]<<" "<<ans[n][1]<<" "<<ans[n][2]<<" "<<ans[n][3]<<" "<<ans[n][4]<<" "<<ans[n][5];  
27.     cout<<" "<<ans[n][6]<<" "<<ans[n][7]<<" "<<ans[n][8]<<" "<<ans[n][9]<<endl;  
28. }  
29.   return 0;  
30. }