UVA 1225 Digit Counting (number of occurrences of the statistical digits) _uva1225

Digit counting
Time limit:3000ms Memory Limit:unknown 64bit IO Format:%lld &%llu

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Description

Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequence to consecutive integers starting with 1 ToN (1 < N < 10000). After this, he counts the number of times each digit (0 to 9) appears in the sequence. For example, withn =, 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 AppE Ars once. After playing for a while, Trung gets bored again. He is now wants to write a the program to the him. The Your task is to help him with writing.

Input

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

For each test case, the there was 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 

Original title Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=27516
The first n (n<=10000) integers written together, such as n=15, 123456789101112131415 to calculate how many times 0-9 each (output 10 numbers, respectively, the number of 0-9 occurrences)
Previously wrote the code, but that is the online code, not very understand, now see an easy to understand the code, on the record down. Previous Blog Links: http://blog.csdn.net/hurmishine/article/details/50880092
AC Code:

#include <iostream>
#include <cstring>
using namespace std;
int main ()
{
    int a[15];
    int t,n;
    cin>>t;
    while (t--)
    {
        memset (a,0,sizeof (a));
        cin>>n;
        for (int i=1;i<=n;i++)
        {
            int t=i;
            while (t)
            {
                int num=t%10;
                a[num]++;
                t/=10
            }
        }
        for (int i=0;i<10;i++)
        {
            if (i)
                cout<< "";
            cout<<a[i];
        }
        cout<<endl;
    }
    return 0;
}