zoj1115 Digital Roots

最新推荐文章于 2019-08-25 15:40:41 发布

neuike

最新推荐文章于 2019-08-25 15:40:41 发布

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本文介绍了一种计算数字根的方法，通过不断累加整数的各位数字直至得到一位数的过程。文章提供了一个C++实现示例，该示例可以处理多达1000位的整数。

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Digital Roots

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 1115

Appoint description:

Description

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

被这道水题给耍了，wa了3次。

不要太天真，这一题整数数位有1000位，只能用字符串存。

#include<cstdio>

#include<cstring>
#include<iostream>

using namespace std;

char ss[10007];
int main()
{
int n;
while(scanf("%s",ss)!=EOF)
{
if(ss[0]=='0')
break;

int tmp=0;
for(int i=0;i<strlen(ss);i++)
{

tmp=tmp+ss[i]-'0';
tmp=tmp%9;

}
tmp=tmp%9;
if(tmp==0)
printf("%d\n",9);
else
printf("%d\n",tmp);

}
return 0;
}