Eddy's digital Roots

最新推荐文章于 2021-08-07 22:56:33 发布

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最新推荐文章于 2021-08-07 22:56:33 发布

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分类专栏： ACM

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本文探讨了如何计算一个整数的幂次后的数字根，深入解析了数字根的概念及计算方法，通过实例展示了求解过程，并提供了有效的编程实现。

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http://acm.hdu.edu.cn/showproblem.php?pid=1163

Eddy's digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5203 Accepted Submission(s): 2910

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

Sample Output

4
4

Author

eddy

提醒一下，因为最后的答案是每一位相加，所以在前面计算的过程中个位和十位是没啥区别的，可以先相加起来在计算也是对的，自己体会。当然这一题也可以同九余数定理。

#include <cstdio>
int fun(int n)
{
    int ans=1;
    for(int i=0;i<n;i++){
        ans*=n;
        while(ans>=10){
        int sum=0;
        while(ans){
            sum+=ans%10;
            ans/=10;
           }
        ans=sum;
        }
    }
    return ans;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        if(n==0)break;
        printf("%d\n",fun(n));
    }
    return 0;
}