uva1225 Digit Counting

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

思路：刚开始用的枚举，后来发现这道题不用枚举也特别简单，就是用个一维数组判断每一位然后让他+1就行。

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n,i,a[11],t;
	scanf("%d",&t);
	while(t--){
		memset(a,0,sizeof(a));
		scanf("%d",&n);
		for(i = 1; i<=n; i++){
			int k = i;
			while(k){
				a[k%10]++;
				k/=10;
			}
		}
		for(i  = 0; i<9; i++){
			printf("%d ",a[i]);
		}
		printf("%d
",a[9]);
	}
	return 0;
}