首页 > 其他 > 详细

hdu 1013

时间:2015-03-18 12:17:20      阅读:194      评论:0      收藏:0      [点我收藏+]

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54655    Accepted Submission(s): 17054


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
 

 

Source
 

 

Recommend
We have carefully selected several similar problems for you:  1017 1018 1048 1016 1019 
 

//题目分析:本题重要的是对大数的境况要考虑到,利用字符数组存储大数
//在判断大数的哪一位结束时利用对位取址。还有就是求大数的根
// d%9是这个大数的根当且仅当余数是0时根为9.
#include<stdio.h>
char a[10000];
int main()
{
int d;
char *p;
while(scanf("%s",a)>0){
p=a;d=0;
while(*p){
d+=*p-‘0‘;
p++;
}
if(d==0) break;
d%=9;
if(d==0) d=9;
printf("%d\n",d);
}
return 0;
}

hdu 1013

原文:http://www.cnblogs.com/18339786086zym/p/4346643.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!