Digital Roots 1013
Digital Roots
时间限制(普通/Java):1000MS/3000MS 运行内存限制:65536KByte
总提交:456 测试通过:162
总提交:456 测试通过:162
描述
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.
输入
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.
输出
For each integer in the input, output its digital root on a separate line of the output.
样例输入
24
39
0
样例输出
6
3
当时数组开辟空间1000 死活通不过,稍大一点就可以
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <cstring> using namespace std; int main() { char num[1005]; int len; while(cin.getline(num,1005)&& num[0]!='0') { len=strlen(num); int sum=0; for(int i=0;i<len;i++) { sum+=num[i]-'0';//一般程序不用 48 而是0’ if(sum>=10) sum=sum/10+sum%10; //这个我开学讲过 如何求解一个数的每一位 } cout<<sum<<endl; } return 0; }
每天明白一点知识