51Nod-1116K进制下的大数(java版本)

题目链接

https://vjudge.net/problem/51Nod-1116

思路

这里用不到大数运算,用的是数学知识
主要就是要理解题意

代码

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class Main {

	/**
	 * @param args
	 * @throws Exception 
	 */
	public static void main(String[] args) throws Exception {
		String s = _08Reader.next();
		int len = s.length();
		int ans = 0;
		int max = 0;
		for (int i = 0; i < len; i++) {
			char charAt = s.charAt(i);
			if (charAt>='0'&& charAt<='9') {
				//转换成数字
				charAt-='0';
			}
			if (charAt>='A'&& charAt<='Z') {
				//转换成数字
				charAt=(char) (charAt - 'A' + 10);
			}
			if (max < charAt) {
				//max=10
				max = charAt;
			}
			//ans=21
			ans += charAt;
		}
		boolean flag = false;
		//从11进制开始
		for (int i = max+1; i <= 36; i++) {
			//这个数在K进制下是K - 1的倍数
			if (ans%(i-1)==0) {
				flag = true;
				System.out.println(i);
				break;
			}
		}
		if (flag == false) {
			System.out.println("No Solution");
		}
	}

}
class _08Reader {
	static BufferedReader br = new BufferedReader(new InputStreamReader(
			System.in));
	static StringTokenizer tokenizer = new StringTokenizer("");

	static String next() throws Exception {
		while (!tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(br.readLine());
		}
		return tokenizer.nextToken();
	}

	static String nextLine() throws Exception {
		return br.readLine();
	}

	static int nextInt() throws Exception {
		return Integer.parseInt(next());
	}

}