HDU1013 Digital Roots求解

maple1116 2013-01-21 08:30:07
题目如下:
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


我的代码如下,交到OJ上去总是wrong answer,请问各路大神这是为什么啊?
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
char a[1000];
int n,i,length,t,b[4];
while(cin.getline(a,1000))
{
if(a[0]=='0')break;
length=strlen(a);
t=a[0]-48;
for(i=1;i<length;i++)
{
t+=a[i]-48;
}
while(t>9)
{
for(i=0;i<4;i++)
{
b[i]=t%10;
t/=10;
}
for(i=0;i<4;i++)
{
t+=b[i];
}
}
cout<<t<<endl;
}
return 0;
}
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maple1116 2013-01-22
引用 2 楼 FancyMouse 的回复:
你得先确认不是因为数组开1000开太小
还真是,改成2000就AC了,谢谢了
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FancyMouse 2013-01-22
你得先确认不是因为数组开1000开太小
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lee_鹿游原 2013-01-21

#include <iostream>
#include <string>
using namespace std;
void DigitRoot(int n)
{
	int tmp;
	int sum = 0;
	while (n!=0)
	{
		tmp = n%10;
		sum += tmp;
		n = n/10;
	}
	if(sum>=0&&sum<=9)
	{
		cout<<sum<<endl;
		return;
	}
	else
	{
		DigitRoot(sum);
	}
}
int main(int argc, char *argv[])
{
	string strNum;
	int i,num;
	while(cin>>strNum&&strNum!="0")
	{
		num = 0;
		int len = strNum.length();
		for (i=0;i<len;++i)
		{
			num += strNum[i]-'0';
		}
		DigitRoot(num);
	}
	return 0;
}
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