1225-Digit Counting, ACM/ICPC,Danang 2007_acm12345678910111213-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_43323172/article/details/86707811

本文介绍了一个编程问题，任务是计算从1到N的连续整数序列中每个数字（0到9）出现的次数。通过构造字符串并遍历字符，统计各数字频率，最终输出结果。

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Time limit: 3.000 seconds

Problem 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

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<ctype.h>
# define maxn 1000000
char str[maxn],s[10];//str用于构造从一到n,s用于存数字
int main()
{
    int d=0,t,n;
    scanf("%d",&t);
    int num[10];
    while(t--){
        memset(str,0,sizeof(str));memset(s,0,sizeof(s));
        memset(num,0,10*sizeof(int));
        scanf("%d",&n);
        for( int i=1; i<=n; i++ ){//构造从一到n的字符串
            sprintf(s,"%d",i);
            strcat(str,s);
        }
        for( int i=0; i<strlen(str); i++ ){//对各个数字计数
            switch(str[i]){
                case'0':num[0]++;break;
                case'1':num[1]++;break;
                case'2':num[2]++;break;
                case'3':num[3]++;break;
                case'4':num[4]++;break;
                case'5':num[5]++;break;
                case'6':num[6]++;break;
                case'7':num[7]++;break;
                case'8':num[8]++;break;
                case'9':num[9]++;break;
                default:break;
            }
        }
        for( int i=0; i<=9; i++ ){
            if(i!=0) printf(" ");
            printf("%d",num[i]);
        }
        printf("\n");
    }
    return 0;
}