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Digital Roots

时间:2015-01-11 11:03:17      阅读:223      评论:0      收藏:0      [点我收藏+]

标签:printf   数据   测试   

Digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52224 Accepted Submission(s): 16315


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 Input
24 39 0

Sample Output
6 3
解题思路:由于没有说明数字位数,并不排除位数很大的数字,所以将输入数据作为字符串处理
错误代码:
#include <stdio.h> #include <stdlib.h> void digits(int n)/*题中没有说明数的位数,所以可能很大,应该是用字符串存储输入数据*/ {   int r,sum = 0;   while(n != 0)   {     r= n%10;     n = n/10;          sum += r;         }   if(sum > 10)     digits(sum);    else     printf("%d\n",sum);  } int main() {   int n;   while(scanf("%d",&n) && n != 0)     digits(n);                        system("pause");   return 0;     }
正确代码:
#include <iostream>    #include <string> using namespace std;    int main()
{            string str;        while(cin>>str&&str!="0")
{    int num=0;                              for(int i=0;i<str.length();i++)
{                 num+=str[i]-‘0‘;              if(num>9)
{                num=num/10+num%10;                        }            }                cout<<num<<endl;                              }           system("pause");            return 0;    }  

Digital Roots

标签:printf   数据   测试   

原文地址:http://blog.csdn.net/zchlww/article/details/42607763

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