UVA 1225 Digit Counting(统计数位出现的次数)

Digit Counting Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu SubmitStatus 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 that, he counts the number of times each digit (0 to 9) appears in the sequence. For example, withN = 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

原题链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=27516

题意:

把前n(n<=10000)个整数顺次写在一起,如n=15时,123456789101112131415

计算0-9各出现了多少次(输出10个数,分别是数字0-9出现的次数)

以前就写了代码,不过那是网上的代码,不是很懂,现在看到了一份容易理解的代码,就记录下来.

以前博客链接:http://blog.youkuaiyun.com/hurmishine/article/details/50880092

AC代码:

#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;
}