POJ 2826 An Easy Problem!（简单数论）

Description

Have you heard the fact “The base of every normal number system is 10” ?

Of course, I am not talking about number systems like Stern Brockot Number System. This problem has nothing to do with this fact but may have some similarity.

You will be given an N based integer number R and you are given the guaranty that R is divisible by (N-1). You will have to print the smallest possible value for N. The range for N is 2 <= N <= 62 and the digit symbols for 62 based number is (0..9 and A..Z and a..z). Similarly, the digit symbols for 61 based number system is (0..9 and A..Z and a..y) and so on.

Input

Each line in the input will contain an integer (as defined in mathematics) number of any integer base (2..62). You will have to determine what is the smallest possible base of that number for the given conditions. No invalid number will be given as input. The largest size of the input file will be 32KB.

Output

If number with such condition is not possible output the line “such number is impossible!” For each line of input there will be only a single line of output. The output will always be in decimal number system.

Sample Input

3

5

A

Sample Output

4

6

11

题意是：对于给定数字。能否使他是N进制，而且满足被N-1整除。假设能找到这样一个数N。那么输出N，否则输出such number is impossible!

首先，举个例 假如这个数是N进制的 2 A C D，这个数值为（2*N*N*N+A*N*N*+C*N+D）,再对N-1取模：

先对第一项2*N*N*N取模，2%（N-1）N%（N-1）*N%（N-1）*N%（N-1）。N%（N-1）=1,所以第一项的值为2%（N-1)，依据这个结果，能够得出（2*N*N*N+A*N*N+C*N+C）%（N-1）=（2+A+C+D)%(N-1)

那么这道题仅仅需把每一位数字的值加起来再对（N-1）取模就可以。

枚举N的值，推断是否取模结果为0。还要对N的范围加以限定。

#include<iostream>
#include<stdio.h>
#include<queue>
#include<stack>
#include<algorithm>
#include<string.h>
#include<string>
#include<math.h>
using namespace std;
int num[30005];
char s[30005];
int main()
{
    while(scanf("%s",&s)!=EOF)
    {
        int len=strlen(s);
        int Max=0;
        for(int i=0;i<len;i++)
        {
            if(s[i]>='0'&&s[i]<='9')
                num[i]=s[i]-'0';
            else if(s[i]>='A'&&s[i]<='Z')
                num[i]=s[i]-'A'+10;
            else if(s[i]>='a'&&s[i]<='z')
                num[i]=s[i]-'a'+36;
            Max=max(Max,num[i]);//比方输入为A，那么进制至少从A開始寻找满足要求的进制
        }
        if(Max==0)
        {
            printf("2\n");//假设输入的数最大为0，那么仅仅能是2进制
        }
        int sum=0;
        for(int i=0;i<len;i++)
        {
            sum+=num[i];
        }
        int flag=100;
        for(int i=Max;i<=62;i++)
        {
            if(sum%i==0)
            {
                flag=i;
                break;
            }
        }
        if(flag<=61)
        {
            printf("%d\n",flag+1);
        }
        else
        {
            printf("such number is impossible!\n");//由于仅仅有62进制。大于61就不满足题目条件了
        }
    }
    return 0;
}