HDU1163 Eddy's digital Roots【快速模幂+九余数定理＋水题】

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 84156    Accepted Submission(s): 26272

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.

 

Input

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.

 

Output

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

 

Sample Input

24 39 0

 

Sample Output

6 3

 

Source

Greater New York 2000

问题链接：HDU1013 Digital Roots。

问题分析：

　　这个问题是对于输入的n，计算n^n的数根。给出两种解题思路如下：

　　解法一：

　　先看一下以下式子：

　　因为：(10*a+b)*(10*a+b)=100*a*a+10*2*a*b+b*b　

　　所以右边式子的数根（中间结果，也是左边式子的数根）为：a*a+2*a*b+b*b=(a+b)*(a+b)

　　故：对于两位数n，n*n的数根=n的树根×n的树根。

　　同理可以推出，对于任意位数的n，也满足：n*n的数根=n的树根×n的树根。　

　　程序中，实现一个计算整数数根的函数，利用这个函数来计算n^n的数根。

　　解法二：

　　计算n^n的数根，一要快，二要简单。使用快速模幂计算，加上数论中的九余数定理就完美了。

AC的C++语言程序如下（解法二）：

/* HDU1163 Eddy's digital Roots */

#include 

using namespace std;

const int NICE = 9;

// 快速模幂计算函数
int powermod(int a, int n, int m)
{
    int res = 1;
    while(n) {
        if(n & 1) {        // n % 2 == 1
            res *= a;
            res %= m;
        }
        a *= a;
        a %= m;
        n >>= 1;
    }
    return res;
}

int main()
{
    int n, ans;

    while(cin >> n && n) {
        ans = powermod(n, n, NICE);
        cout << (ans ? ans : 9) << endl;
    }

    return 0;
}

　

AC的C语言程序如下（解法一）：

/* HDU1163 Eddy's digital Roots */

#include 

// 计算数根函数
int digitalroots(int val)
{
    int result, temp;

    while(val) {
        result = 0;
        temp = val;

        while(temp) {
            result += temp % 10;
            temp /= 10;
        }

        if(result < 10)
            break;

        val = result;
    }

    return result;
}

int main(void)
{
    int n, ans, nr, i;

    while(scanf("%d", &n) != EOF) {
        if(n == 0)
            break;

        // 计算n的数根
        ans = nr = digitalroots(n);

        // 计算n^n的数根
        for(i=2; i<=n; i++) {
            ans = digitalroots(ans * nr);
        }

        // 输出结果
        printf("%d\n", ans);
    }

    return 0;
}