Eddy's digital Roots

时间：2015-02-01 12:00:27 阅读：180 评论：0 收藏：0 [点我收藏+]

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4793 Accepted Submission(s): 2672

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

思路：1.九余数定理一个数的n的树根=（n-1)%9+1

2.N个整数的树根的乘机总值=每个项元素的树根的乘积

3.分步求余

#include<cstdio>
#include<cmath>
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        int m=n;
        int i;
        int ans=1;
        m=(m-1)%9+1;//先把底数m求树根，根据  N个整数的乘积带的树根=每项元素的树根的乘积
        for(i=1;i<=n;i++)
        ans=(ans*m-1)%9+1;//分步求余
        printf("%d\n",ans);
    }
}

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Eddy's digital Roots

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原文地址：http://www.cnblogs.com/orchidzjl/p/4265334.html

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