Uva 1225 - Digit Counting-优快云博客

本文链接：https://blog.youkuaiyun.com/Coder_Jeremy/article/details/52027862

本文介绍了一个简单的程序设计问题，即统计从1到给定整数N中每个数字0至9出现的次数。通过使用C语言实现，采用桶排序的方法来统计各个数字出现的频率，并提供了解决方案的完整代码。

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1225 - 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

题意很好理解，输入数字n，统计从1到n中每个数字出现的次数
用桶排法可以很简单解决

#include <cstdio>
#include <cstring>

int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int a;
        int cnt[10]={0};
        int x;
        scanf("%d",&a);
        int m;
        for(int i=1; i<=a; i++)
        {
            m=i;
            while(m!=0)
            {
                x=m%10;
                cnt[x]++;
                m=m/10;
            }
        }
        for(int i=0; i<10; i++)
        {
            printf("%d",cnt[i]);
            if(i!=9)printf(" ");
        }
        printf("\n");
    }
    return 0;
}