POJ 1503 Integer Inquiry(大数相加)

一、Description

One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.


``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.)

Input

The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative).




The final input line will contain a single zero on a line by itself.

Output

Your program should output the sum of the VeryLongIntegers given in the input.

二、题解

        这个题和之前2602有Poj 2602 Superlong sums(大数相加)异曲同工。但还是有要注意的部分，就是这里有前导的0，而且题目没说是否每个数的位数是否相同。还有数组的长度也比题目说明的要大，所以开数组的时候千万不要手下留情啊。这个题目我WA了还多次，可能是考虑的情况不够全面，对题目的要求没有完全弄明白。所以，虽然经过了几次的调试，几组测试数据都过了，但是还是没有AC。请大神指教。但当我一头雾水的时候呢，居然发现java有一种脑残解法，由于java 中自带的 BigInteger 类，这是个不可变的任意精度的整数。而且接受十进制的字符串，并能将 BigInteger
的十进制字符串表示形式转换为 BigInteger。所以，管它什么大数，都弱爆了。但是看来好东西虽然存在，但是了解它的原理还是必要的。

三、java代码

import java.util.*;
import java.math.*;   

public class Main {   

    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        BigDecimal bd1 = BigDecimal.valueOf(0);
        BigDecimal bd2 = BigDecimal.valueOf(0);
        String str;   

        while(cin.hasNext()){
            str = cin.nextLine();
            if(str.equals("0"))
                break;
            else{
                bd2 = new BigDecimal(str);
                bd1 = bd1.add(bd2);
            }
        }   

        System.out.println(bd1.toPlainString());
    }
}  

求错误代码！！

import java.io.IOException;
import java.util.Scanner;

  public class Main {
	  static byte[] c = new byte[1000];
      public static void add(String s1){
    	  int i;
          for (i = 0; i <s1.length(); i++){
              c[i] += s1.charAt(i) - 48;
          }
          int cf = 0;
          for (i = s1.length() - 1; i >= 0; i--) {
              c[i] += cf;
              cf=c[i]/10;
              c[i]=(byte) (c[i] %10);
          }
      }
	  public static void main(String[] args) throws IOException {
       Scanner cin = new Scanner(System.in);
       String s[]=new String[120];
       int i=0,j=0,k;
       String ss,te;
       String te2;
       int max=0;
       while(!(ss=cin.next()).equals("0")){
    	   s[i]=ss;
    	   max=Math.max(s[i].length(), max);
    	   i++;
       }
       for(k=0;k<i;k++){
    	   te=s[k];
    	   te2 ="";
    	   while(s[k].length()<max){
    		   s[k]+="0";
    		   te2+="0";
    	   }
    	   s[k]=te2+te;
    	   add(s[k]);
       }
       while(c[j]==0){
	    	j++;
	    }
       for(k=j;k<max;k++){
    	    System.out.print(c[k]);
       }
       System.out.println();
    }
 }     

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