11332 - Summing Digits
For a positive integer n
, let f(n)
denote the sum of the digits of n
when represented in base 10. It is easy to see that the sequence of numbers n, f(n), f(f(n)), f(f(f(n))), ...
eventually becomes a single digit number that repeats forever. Let this single digit be denoted g(n)
.
For example, consider n = 1234567892
. Then:
f(n) = 1+2+3+4+5+6+7+8+9+2 = 47 f(f(n)) = 4+7 = 11 f(f(f(n))) = 1+1 = 2
Therefore, g(1234567892) = 2
.
Each line of input contains a single positive integer n
at most 2,000,000,000. For each such integer, you are to output a single line containing g(n)
. Input is terminated by n = 0
which should not be processed.
对一个正整数n, f(n) 表示n 以十进位表示时所有位数的和。显而易见地,数列n, f(n), f(f(n)), f(f(f(n))), ...一直重覆下去最后会变成一个一位数的整数。这个一位数字以 g(n) 表示。
例如,若 n = 1234567892,则:
f(n) = 1+2+3+4+5+6+7+8+9+2 = 47
f(f(n)) = 4+7 = 11
f(f(f(n))) = 1+1 = 2
因此, g(1234567892) = 2。
Sample input
2 11 47 1234567892 0
Output for sample input
2 2 2 2
解题思路:把给定的数拆开来,相加如果大于9,再拆再加直到和小于10为止
#include<stdio.h> long f(long n) {long sum=0,a; while(n){ sum+=n%10; n/=10; } if(sum>=10)a=f(sum); else a=sum; return a; } int main() {long n; while(scanf("%ld",&n)!=EOF){ if(n==0)break; printf("%ld\n",f(n)); } return 0; }