POJ-1152 An Easy Problem! 解题报告（数论） 是不是N进制数

连接----An Easy Problem!

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7802		Accepted: 1884

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

Source
uva 10093


题目大意：给定一个数，看能否用N（2<=N<=62）进制表示，输出能表示的最大进制。（32bit*1024约30000所以开大点的数组）

#include <iostream>
#include<cstring>
using namespace std;
char s[50000];	
int data[200];	
int main()
{
	int i;
	for(i='0';i<='9';i++)
		data[i]=i-'0';
	for(i='A';i<='Z';i++)
		data[i]=i-'A'+10;
	for(i='a';i<='z';i++)
		data[i]=i-'a'+36;//data[ASCII码]=1~62的数字，建立十进制库

	while(cin >> s)	//这个用scanf还“output limit exceeded”了！
	{
		int len=strlen(s);
		int sum=0,max=0,result=0;
		for(int j=len-1;j>=0;j--)
		{	sum+=data[(int)s[j]];//s%(n-1)=(a *n^3 % (n-1) + b *n^2 % (n-1) + c *n % (n-1) + d % (n-1)) % (n-1)
		//再想想，n^k % (n-1)应该等于1，直接去掉，只剩 s %(n-1)= sum % (n-1);
			if(data[(int)s[j]]>max) max=data[(int)s[j]];
		}
		for(i=max+1;i<=62;i++)
			if(sum%(i-1)==0) {result=i;break;}
		if(result==0) printf("such number is impossible!\n");
		else printf("%d\n",result);
	}
	return 0;
}