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主题:有一道算法题,不知各位能否帮我看看错在哪里?

Background

The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.


Input

The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.


Output

For each integer in the input, output its digital root on a separate line of the output.


Example

Input



24
39
0
Output

6
3

意思就是说算一个正整数各位的和,如果答案不是一位数,就继续算下去,直到结果是一位数为止。

以下是我写的代码(C++)
#include<iostream.h>

int digit(int &num)
{
        int result = num;
        while(result>9)
        {
            result = 0;
            while(num>9)
            {
                result += num%10;
                num = num/10;
            }
            result += num;
            num = result;
        }
        return result;
}


int main()
{
    int n,i;
    int *data = new int[100];
    int *result = new int[100];

    cout<<"Please input several positive integers, one per line, \nended by an integer value of zero:"<<endl;

    for(i=0;;i++)
    {
        cin>>data[i];
        if(!data[i])
        {
            n=i;
            break;
        }
    }

    cout<<"The output: their digital roots:"<<endl;

    for(i=0; i<n; i++)
    {
        result[i] = digit(data[i]);
        cout<<result[i]<<endl;
    }

    delete []data;
    delete []result;
    return 0;
}


自己觉得没什么错误,但交到在线答题系统里,说我的是wrong answer,为什么啊?

[em7]

回复列表 (共3个回复)

沙发

怎么没有人回复啊。。。

板凳

在你的程序的第四行,为什么要result 小于 9 才结束循环?我看了一下好像是这里有问题.

3 楼


要用字符型的,要不数字大了就不行了
这是我在网上找到的你可以看一下,用C语言写的
#include <stdio.h>
#include <string.h>

int main()
{
 char s[1000],*p;
 int n;
 while(scanf("%s",s)!=EOF){
  if(strcmp(s,"0")==0)
   break;
 while(strlen(s)>1){
   for(n=0,p=s;*p;p++)
    n+=*p-'0';
   sprintf(s,"%d",n);
  }
  printf("%s\n",s);
 }
}

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