UVa 1225 Digit Counting 题解

英文

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

中文

题目大意：

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)

提示：

写一个子程序来累加一个数中每个数字的出现个数就行了。小心UVa中莫名其妙的格式错误。

代码：

#include
#include
#include
#include
#include
#include
using namespace std;
long long a[11],n,T;
void pro(long long k)
{
    while(k>0)
    {
        a[k%10]++;
        k/=10;
    }
}
int main()
{
    scanf("%lld",&T);
    for(int i=1;i<=T;i++)
    {
        memset(a,0,sizeof(a));
        scanf("%lld",&n);
        for(int j=1;j<=n;j++) pro(j);
        for(int j=0;j<=8;j++) printf("%lld ",a[j]);
        printf("%lld\n",a[9]);
    }
    return 0;
}

相关链接：

UVa题解小全：
https://blog.csdn.net/zj_mrz/article/details/81144019

UVa 1583 Digit Generator 生成元 题解：
https://blog.csdn.net/zj_mrz/article/details/81143855

UVa 1585 Score 得分 题解：
https://blog.csdn.net/zj_mrz/article/details/81144159

转载于:https://www.cnblogs.com/zj-mrz/p/10122457.html