HDOJ1013 Digital Roots（高精度+数位分离求和）

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 80317    Accepted Submission(s): 25127

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

24390

 

Sample Output

63

 

Source

Greater New York 2000

【分析】数根问题：计算正整数n的各位数字之和（记为sum）。

        （1）若n是个位数，则n的数根是其本身；

        （2）若求得的sum是个位数，则n的数根为sum；

        （3）否则对sum求各位数字之和，直到结果为个位数为止。

        这里需要注意n是大整数，因此用字符数组（字符串）保存之。

#include <stdio.h>#include <string.h>char num[1010];int main(){int i,n;int sum,temp;    while(scanf("%s",num)!=EOF)    {    sum=0;    if(strcmp(num,"0")==0)    break;    for(i=0;i<strlen(num);i++)    sum+=(num[i]-'0');    n=sum;while(n>=10)    {    temp=n;    sum=0;    while(temp!=0)    {    sum+=(temp%10);    temp/=10;}n=sum;}printf("%d\n",n);}    return 0;}