Digital Roots

最新推荐文章于 2020-05-10 23:42:17 发布

原创最新推荐文章于 2020-05-10 23:42:17 发布 · 148 阅读

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CC 4.0 BY-SA版权

本文探讨了如何通过编程计算正整数的数字根，即不断累加整数的各位数字直至得到一位数的过程。文章提供了两种解决方案：一种适用于较小数值的直接整数操作；另一种则使用字符串处理，能有效处理大数值，展示了字符串在处理大整数问题上的优势。

Total Submission(s) : 44 Accepted Submission(s) : 11

Problem Description 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.

Sample Input24
39
0

Sample Output6
3

#include<stdio.h>
#include<string.h>
int sum(int n);
int main()
{
 char a[10000];
 while(gets(a)&&strcmp(a,"0"))
 {
  int s=0;
  for(int i=0;a[i]!='\0';i++)
   s+=a[i]-'0';
  while(s>=10)
  {
   s=sum(s);
  }
  printf("%d\n",s);
 } 
 return 0;
} 
int sum(int n)
{
 int a=0;
 while(n>0)
 {
  a+=n%10;
  n/=10;
 }
 return a; 
}

先把我的错误代码也发一下吧，这个代码读不了太大的值，最后请教了一下同学，这个题需要用字符串（没有表情来表达我内心的想法）

#include<stdio.h>
int sum(int n);
int main()
{
 int n;
 while(scanf("%d",&n)&&n>0)
 {
   
  while(n>=10)
  {
   n=sum(n);
  }
  printf("%d\n",n);
 } 
 return 0;
} 
int sum(int n)
{
 int a=0;
 while(n>0)
 {
  a+=n%10;
  n/=10;
 }
 return a; 
}