Digit Counting UVA - 1225

最新推荐文章于 2022-09-09 23:31:54 发布

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最新推荐文章于 2022-09-09 23:31:54 发布

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原文链接：http://www.cnblogs.com/yongzi/p/7363051.html

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，并通过示例代码展示了如何实现这一功能。

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.

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

这个题目刚做的时候觉得非常简单，一会就敲完啦，但交上去之后显示时间超时，问题出现在输入的数字上，在逻辑上面，我的程序已经出现了漏洞。不多说了，看代码吧；

#include<stdio.h>
#include<string.h>
#define maxn 10
int main()
{
    int N;
    int a[maxn];
    scanf("%d",&N);
    while(N--){
        int n,i,j;
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(i=1;i<=n;i++){
             //for(j=i;j>0;j/=10)
               // a[j%10]++;
               j=i;
               while(j)
               {
                   a[j%10]++;
                   j/=10;
               }
        }
        printf("%d",a[0]);
        for(i=1;i<10;i++)
            printf(" %d",a[i]);
        printf("\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/yongzi/p/7363051.html