HDOJ 1013 Digital Roots（大数）

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 58524    Accepted Submission(s): 18278

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 Greater New York 2000  

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就是一个大数求和问题，我竟然。。。

第一次用整型，WA了

第二次没初始化，WA了

第三次AC:

#include<stdio.h>
#include<string.h>
int main(){
	char a[10000];
	int la,b[10000],sum,i;
	while(scanf("%s",a)&&strcmp(a,"0")!=0){
		la=strlen(a);
		sum=0;//各位数和 
		for(i=0;i<la;i++){
			b[i]=a[i]-'0';//字符转数字 
			sum+=b[i];//一个个求和 
			if(sum>=10)//超过10，最多18 
			sum=sum%10+sum/10;//两位数再加起来 
		}
		printf("%d\n",sum);
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
	}
	return 0;
}
           

根数：公式为b=(a-1)%9+1

java:

import java.io.*;
import java.math.*;
import java.util.*;
public class big2 {
	public static void main(String[] args){
		Scanner input=new Scanner(System.in);
		while(input.hasNext()){
			String num=input.next();
			if(num.equals("0"))
				break;
			BigInteger big=new BigInteger(num);
			BigInteger sum=big.add(new BigInteger("-1"));
			sum=sum.mod(new BigInteger("9"));
			sum=sum.add(BigInteger.ONE);
			System.out.println(sum);
		}
	}
}