12^(9^(9^(9^(9^(9^(9^(9^(9^9))))))))

12^(9^(9^(9^(9^(9^(9^(9^(9^9))))))))

#include <time.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include "../../../HugeCalc_API/CppAPI/Include/HugeCalc.h"  // 公共接口
#include "../../../HugeCalc_API/CppAPI/Include/HugeInt.h"   // 10进制系统
#include "../../../HugeCalc_API/CppAPI/Include/HugeIntX.h"  // 16进制系统

#pragma message( "automatic link to ../../../HugeCalc_API/CppAPI/Lib/HugeCalc.lib" )
#pragma comment( lib, "../../../HugeCalc_API/CppAPI/Lib/HugeCalc.lib" )

#define integer CHugeInt

#define MAX 1000
int sum_of_digits(integer u)
{
    int sum=0;
    integer d(u);
    while(d!=0){
        sum+=d%10;
        d/=10;
    }
    return sum;
}
int main()
{
    int i,j;
    srand(time(NULL));
    for(i=0;i<MAX;i++){
        integer x;
        x.Random(77);
        x+=1-x%3;
        for(j=0;j<MAX;j++){
            integer y(x*x);
            if(sum_of_digits(y)==772){
                printf("%s, %d, %s\n",x.GetStr(FS_NORMAL),x%3,y.GetStrA(FS_NORMAL));
                return 0;
            }
            x+=3;
        }
    }
}

public class Guess {

    /**
     * 任何一个不能被3整除的偶数，如488，按下列步骤：
     * 若该数为偶数，则把它各个位数之和的平方作为新数；
     * 若该数为奇数则各个位数之和的立方作为新数，
     * 再把那个新数重复以上步骤(偶数就各位数之和平方,奇数就各位数之和立方)，
     * 一步步计算下去,肯定能在9步内变为1!
     *
     * @param args -
     */
    public static void main(String[] args) {
        for (int i = 2; i < 10000; i+= 2) {
            if (i % 3 != 0) {
                process(i);
            }
        }
    }

    private static void process(int n) {
        int original = n;
        int step = 0;

        while (step < 10 && n > 1) {
            int m = sum(n);
            if (n % 2 == 0) {
                n = m * m;
            } else {
                n = m * m * m;
            }
            step++;
        }

        if (step == 10) {
            System.out.println(original + " is invalid.");
        }
    }

    private static int sum(int n) {
        int sum = 0;
        while(n >= 10) {
            sum += n % 10;
            n = n / 10;
        }
        return sum + n;
    }
}