hdu1163 Eddy's digital Roots

最新推荐文章于 2020-07-21 19:11:21 发布

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本文介绍了一种计算数字根的方法，通过不断累加数字直至得到一位数的过程，并给出了一种快速求解n^n数字根的算法实现，利用取模运算避免了大数运算的复杂性。

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.

The Eddy's easy problem is that : give you the n,want you to find the n^n's digital Roots.

Input

The input file will contain a list of positive integers n, one per line. The end of the input will be indicated by an integer value of zero. Notice:For each integer in the input n(n<10000).

Output

Output n^n's digital root on a separate line of the output.

Sample Input

2
4
0

Sample Output

4
4

解答：

这答题就是常见的求数的根。

比如39，就是3+9=12，但是12>9,所以还得继续求，1+2=3，所以39的根是3.

对于所有的数的根都是这样，一直求到个位数为止。

求数的根有个规律就是，一个数的根等于这个数%9之后的数，比如%9后是0，就说明这个数根是9.其他的就是原数

本来以为这道题求n次方得用快速幂，可是看题解没用快速幂也没有超时。

贴代码：

#include"iostream"
using namespace std;
int main()
{
    int n,a,sum,i;
    while(cin>>n&&n)
    {
        sum=1;
        for(i=0;i<n;i++)
        {
            sum=sum*n%9;
        }
        if(sum==0)
        cout<<"9"<<endl;
        else
        cout<<sum<<endl;
    }
    return 0;
}