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题意:a = 1, b = 2, ..., z = 26, aa = 27, ...
给字符串或者数字,输出对应的答案。
解法:类似26进制……但又不完全是……拿java大数模拟了一下……
代码:
import java.util.*; import java.math.*; public class Main { public static void main(String args[]) { Scanner cin = new Scanner(System.in); while(true) { String input = cin.next(); if(input.equals("*")) break; String ans1 = "", ans2 = ""; if(input.charAt(0) >= ‘0‘ && input.charAt(0) <= ‘9‘) { ans2 = input; BigInteger x = new BigInteger(input); while(!x.equals(BigInteger.ZERO)) { if(x.equals(BigInteger.valueOf(26))) { ans1 += ‘z‘; break; } ans1 += (char)(‘a‘ + (Integer.parseInt(x.mod(BigInteger.valueOf(26)).toString()) + 25) % 26); if(x.mod(BigInteger.valueOf(26)).equals(BigInteger.ZERO)) { x = x.divide(BigInteger.valueOf(26)); x = x.add(BigInteger.valueOf(-1)); } else x = x.divide(BigInteger.valueOf(26)); } } else { ans1 = input; BigInteger x = BigInteger.ZERO; for(int i = 0; i < input.length(); i++) { x = x.multiply(BigInteger.valueOf(26)); x = x.add(BigInteger.valueOf((int)input.charAt(i) - ‘a‘ + 1)); } ans2 = x.toString(); } if(input.charAt(0) >= ‘0‘ && input.charAt(0) <= ‘9‘) for(int i = ans1.length() - 1; i >= 0; i--) System.out.print(ans1.charAt(i)); else System.out.print(ans1); for(int i = 22 - ans1.length(); i > 0; i--) System.out.print(" "); int len = ans2.length() % 3; int res = ans2.length(); for(int i = 0; i < len; i++) System.out.print(ans2.charAt(i)); for(int i = len; i < res; i += 3) { if(i != 0) System.out.print(‘,‘); System.out.print(ans2.charAt(i)); System.out.print(ans2.charAt(i + 1)); System.out.print(ans2.charAt(i + 2)); } System.out.println(""); } } }
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原文地址:http://www.cnblogs.com/Apro/p/4776738.html