Digit Counting

计数序列中数字出现次数

最新推荐文章于 2021-12-18 15:35:54 发布

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本文介绍了一个程序设计问题，即如何计算从1到N的连续整数序列中每个数字0-9出现的频率，并提供了完整的C语言实现代码。

Time limit: 3.000 seconds

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 digit0, 1,...9 separated by a space.

译：把前n（n<=10000）个整数顺次写在一起；123456789101112……数一数0~9各出现多少次（输出10个整数，分别是0，1，2，3，……，9出现的次数）。

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

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int num_Nums[11];                                
    int n, i, j, k, num;
    scanf("%d", &n);
    for(i = 0; i < n; i++)
    {
        memset(num_Nums, 0, sizeof(num_Nums));          //initialize for each loop
        scanf("%d", &num);
        for(j = 1; j <= num; j++){
            int a, b;                                   //a records the result of mod, b avoid the change of j
            b = j;
            while(b > 0){
                a = b % 10;
                num_Nums[a]++;
                b /= 10;
            }
        }
        for(k = 0; k < 10; k++)
        printf("%d ", num_Nums[k]);
        printf("\n");
    }
    return 0;
}