UVA 1225 Digit Counting (number of statistics occurrences)

Digit Counting Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld &%llu Submit Status 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 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 all digit from 4 to 9 AppE Ars once. After playing for a while, Trung gets bored again. He now wants to the write a program to does this for him. The Your task is to the help him with the 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 was a positive integer and is not bigger than 20. The following lines describe the data sets. For each test case, there was one single line containing the number N . Output For each test case, write sequentially on 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
Test Instructions:writes the first n (n<=10000) integers sequentially, such as n=15, 123456789101112131415calculate how many times each 0-9 appears (output 10 numbers, respectively, number 0-9 occurrences)
previously wrote the code, but that is the online code, not very understand, now see an easy to understand the code, it is recorded.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;}

UVA 1225 Digit Counting (number of statistics occurrences)