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2013
11-11

POJ 2602 Superlong sums [解题报告] Java

Superlong sums

问题描述 :

The creators of a new programming language D++ have found out that whatever limit for SuperLongInt type they make, sometimes programmers need to operate even larger numbers. A limit of 1000 digits is so small… You have to find the sum of two numbers with maximal size of 1.000.000 digits.

输入:

The first line of an input file contains a single number N (1<=N<=1000000) - the length of the integers (in order to make their lengths equal, some leading zeroes can be added). It is followed by these integers written in columns. That is, the next N lines contain two digits each, divided by a space. Each of the two given integers is not less than 1, and the length of their sum does not exceed N.

输出:

Output file should contain exactly N digits in a single line representing the sum of these two integers.

样例输入:

4
0 4
4 2
6 8
3 7

样例输出:

4750

温馨提示:

Huge input,scanf is recommended.

解题代码:

import java.io.BufferedInputStream;
 import java.io.BufferedReader;
 import java.io.IOException;
 import java.io.InputStreamReader;

 public class Main {

     public static void main(String[] args) throws NumberFormatException,
             IOException {
         BufferedInputStream read = new BufferedInputStream(System.in);
         byte[] b = new byte[5000007];
         read.read(b);
         String s = "";
         int index = 0;
         while (b[index] != '\r') {
             s += (char) b[index];
             index++;
         }
         int n = Integer.parseInt(s);
         byte[] c = new byte[n];
         for (int i = 0; i < n; i++) {
             index++;
             index++;
             c[i] += b[index] - 48;
             index++;
             index++;
             c[i] += b[index] - 48;
             index++;
         }
         int cf = 0;
         for (int i = n - 1; i >= 0; i--) {
             c[i] += cf;
             if (c[i] >= 10) {
                 cf = 1;
                 c[i] += 38;
             } else {
                 cf = 0;
                 c[i] += 48;
             }
         }
         System.out.write(c);
         System.out.println();
     } 
}