【习题3-3】 UVA - 1225 Digit Counting

最新推荐文章于 2020-09-16 21:16:16 发布

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最新推荐文章于 2020-09-16 21:16:16 发布

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分类专栏：【大一】【紫书】文章标签： UVA 紫书

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本文介绍了一个程序设计问题，即计算从1到N的连续整数序列中每个数字0到9出现的次数。通过枚举的方式实现，并提供了一个C++示例代码。

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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

由于n的范围不大，所以直接枚举即可。。

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int main()
{
    int n, T, i, temp;
    int num[20];
    cin >> T;
    while(T--)
    {
        cin >> n;
        memset(num, 0, sizeof(num));
        for(i = 1;i <= n;i++)
        {
            temp = i;
            while(temp)
            {
                num[temp%10]++;
                temp /= 10;
            }
        }
        for(i = 0;i < 9;i++)
            cout << num[i] << ' ';
        cout << num[i] << endl;
    }
    return 0;
}