POJ 1312 Numerically Speaking

標籤：

題意：a = 1, b = 2, ..., z = 26, aa = 27, ...

給字串或者數字，輸出對應的答案。

 

解法：類似26進位……但又不完全是……拿java大數類比了一下……

 

代碼：

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));}elsex = 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));elseSystem.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("");}}}

　　

POJ 1312 Numerically Speaking