poj-1152 a easy problem

                                                                                   I -An Easy Problem!

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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

题目的大意就是输入一个数，然输出转为新的进制的数，其实主要思路和以前做的转为任意进制是一样的。判断每个数在这个进制中的第几位。只不过这次的数比较特殊罢了，

是由0~9，A~Z,a~z，转化为2~62的数。

#include<stdio.h>
#include<string.h>	
char a[100000];
int main()
{
	int n,i,k,max,b;
	while(scanf("%s",a)!=EOF)
	{
		max=0;
		n=0;
		k=strlen(a);
		for(i=0;i<k;i++)
		{
			if(a[i]>='0'&&a[i]<='9')
			{
				if(a[i]-'0'>max)
				{
					max=a[i]-'0';
				}
				n=n+a[i]-'0';
			}
			if(a[i]>='a'&&a[i]<='z')
			{
				if(a[i]-'a'+36>max)
				{
					max=a[i]-'a'+36;
				}
				n=n+a[i]-'a'+36;
			}
			if(a[i]>='A'&&a[i]<='Z')
			{
				if(a[i]-'A'+10>max)
				{
					max=a[i]-'A'+10;
				}
				n=n+a[i]-'A'+10;
			}
		}
		b=-1;
		for(i=2;i<=62;i++)
		{
			if(n%(i-1)==0&&max<i)
			{
				b=1;
				break;
			}
		}
		if(b==1)
			printf("%d\n",i);
		else
			printf("such number is impossible!\n"); 
	}
	return 0;
}