hdoj1013_Digital Roots

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
开始的时候没有考虑大数的问题，后面知道测试数据有大数的时候注意下就可以了
#include<iostream>#include<string>using namespace std;short num[10000], tnum[10000];//num存储每一次计算的结果，tnum存储每一次计算的中间结果，数组每一位存储计算结果的一位数字string input;int pos = 0;//当前中间计算结果tnum共有几位,也作为每一次计算结果的位数进行参数传递void solve(int n){pos = 0;if (n==0)//如果计算结果中只有一位数字时，输出结果，结束计算cout << num[0] << endl;else{for (int i = 0; i <= n;i++)//每一数据位数字累加{tnum[0] += num[i];for (int j = 0; j <= pos;j++)//判断是否有进位{if (tnum[j] >= 10){tnum[j + 1] += tnum[j] / 10;tnum[j] = tnum[j] % 10;}elsebreak;}if (tnum[pos + 1] > 0)//计算结果位数又增加一位{pos++;tnum[pos+1]=tnum[pos]/10;tnum[pos] %= 10;}}memcpy(num, tnum, sizeof(num));memset(tnum, 0, sizeof(tnum));solve(pos);//再次递归调用}}int main(int argc, char* argv[]){while (cin >> input&&input[0] != '0'){memset(num, 0, sizeof(num));memset(tnum, 0, sizeof(tnum));for (int i = 0; i < input.size(); i++)num[i] = input[i] - '0';solve(input.size()-1);}return 0;}