A well-known trick to know if an integer N is a multiple of nine is to compute the sum S of its digits. If S is a multiple of nine, then so is N. This is a recursive test, and the depth of the recursion needed to obtain the answer on N is called the 9-degree of N.
Your job is, given a positive number N, determine if it is a multiple of nine and, if it is, its 9-degree.
Input
The input is a file such that each line contains a positive number. A line containing the number 0 is the end of the input. The given numbers can contain up to 1000 digits.
Output
The output of the program shall indicate, for each input number, if it is a multiple of nine, and in case it is, the value of its nine-degree. See the sample output for an example of the expected formatting of the output.
Sample Input
999999999999999999999
9
9999999999999999999999999999998
0
Sample Output
999999999999999999999 is a multiple of 9 and has 9-degree 3.
9 is a multiple of 9 and has 9-degree 1.
9999999999999999999999999999998 is not a multiple of 9.
题链接：UVA10922 2 the 9s
问题简述：（略）
问题分析：
输入的整数很长，可能达到1000位，问能否被9整除。如果能则将各位数字相加，直到变成9，计算这个过程的步数并输出结果。
这个问题包含一个大数模除求余数问题，可以一边将数字字符串转换为10进制数一边进行模除取余数。
程序说明：（略）
题记：（略）
参考链接：（略）
AC的C++语言程序如下：
/* UVA10922 2 the 9s */ #include <bits/stdc++.h> using namespace std; const int NINE = 9; const int N = 1000; char s[N + 1]; int main() { while(~scanf("%s", s) && (s[0] != '0' || s[1] != '\0')) { int r = 0, sum = 0; for(int i = 0; s[i]; i++) { sum += s[i] - '0'; r *= 10; r += s[i] - '0'; r %= NINE; } if(r) printf("%s is not a multiple of 9.\n", s); else { int dreep = 1; while(sum > 9) { dreep++; int t = sum; sum = 0; while(t) { sum += t % 10; t /= 10; } } printf("%s is a multiple of 9 and has 9-degree %d.\n", s, dreep); } } return 0; }
