【数论】 大数的各个位数和直到取到个位

2012年3月11日dlut周赛1002

LOVELY-POINT

TimeLimit: 1 Second MemoryLimit: 32 Megabyte

Totalsubmit: 154 Accepted: 42

Description

Lolihunter loves Lolita,To tell while lolita is batter,he makes the lovely-point of a Lolita which is found by summing the digits of the fascination of the Lolita. If the resulting value is a single digit then that digit is the Lolita’s lovely-point. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as much as necessary times to obtain a single digit.
The fascination of a Lolita is n^n.n is a positive integer which is her age.

For example, consider the her age n=4. 4*4*4*4=256. Adding the 2 ，the 5,and the 6 yields a value of 13. Since 13 is not a single digit, the process must be repeated. Adding the 1 and the 3 yeilds 4, the lovely-point of the Lolita is 4.

Input

The input file will contain a list of positive integers, one per line(0<=n<=10000). The end of the input will be indicated by an integer value of zero.

Output

For each integer in the input, output its digital sum on a separate line of the output.

Sample Input

4
3
0

Sample Output

4
9

 

 

 

接触到的第一道数论题，开始以为是大数运算问题，其实可以步步求余来简化。

那么本题的关键在于一个%9

我们看这样的例子abc-->表示为100*a +10*b +c  那么进行各个位的数值相加运算后是什么结果呢？

我们认为是（100*a +10*b +c ）%9的余数，因为（100*a +10*b +c ）可以写成（99*a +9*b）+（a+b+c），推广之~那么我们步步对9求余，如果能被9整除则结果为9，否则继续求余。

 

#include <iostream>

#include <cstring>

#include <cstdio>

#include <stack>

using namespace std;

int main()

{

    //freopen("in.txt","r",stdin);

    int n;

    while(cin >> n)

    {

        int flag=0;

        if (n==0)break;

        int temp=n;

        for(int i=0;i<temp-1;i++)

        {

            if(n%9==0)

            {

                flag=1;

                break;

            }

            n%=9;

            n=n*temp;

        }

        if(flag==1)cout << 9 << endl;

        else

        {

            if(n>9)n%=9;

            cout << n << endl;

        }

    }

    return 0;

}