北大ACM poj1519 Digital Roots

Digital Roots

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 23258 Accepted: 7704

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

 

考虑一下两位数的数字根时，会发现最大的两位数99是：

9+9=18

1+8=9

如果是98，则是：

9+8=17

1+7=8如果还没看出来，就继续用97,96...试试。在计算数字根之后，

如果变成两位数，只要再减去9就是最终数字根了。即使位数增加，

只要考虑反复两位数的计算，就应该能用同样的规则计算数字根。

#include<stdio.h>main(){int ans=0,n;for(;n=getchar();){//首行是0就结束if(ans==0&&n=='0') break;//看到换行符就输出if(n=='\n'){printf("%d\n",ans);ans=0;} else{ans+=n%'0';//ans加上刚刚输入的数（ASCII转化为数） if(ans>9)ans-=9;//如果ans比9大，就减9 }}}