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

最新推荐文章于 2021-02-07 10:15:36 发布

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本文介绍了一种算法，用于解决给定一个整数，在特定条件下寻找能够表示该数的最小可能进制的问题。通过输入一个整数，算法将输出满足条件的最小进制数，范围限制在2到62之间。

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;
}