HDOJ1002

最新推荐文章于 2019-03-30 23:29:13 发布

我只想叫VEE

最新推荐文章于 2019-03-30 23:29:13 发布

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分类专栏： Java 文章标签： HDOJ

本文链接：https://blog.youkuaiyun.com/weixin_40989748/article/details/80420208

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本文介绍了解决HDOJ1002大数相加问题的过程，从最初尝试使用传统方法到利用Java的BigInteger类实现高效解决方案。文中详细解释了两种方法的具体实现，并对比了其优劣。

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HDOJ1002大数相加

问题
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
输入
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
输出
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

用以前C/C++的思维，我第一个想到的是这个

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner =new Scanner(System.in);
        int T=scanner.nextInt();
        for(int i=0;i<T;i++) {
            String A=scanner.next();
            String B=scanner.next();
            int lenA=A.length();
            int lenB=B.length();
            int lenMax=lenA>lenB ? lenA : lenB;
            char[] C=new char[lenMax];
            int lenAA=lenA;
            int lenBB=lenB;
            int before=0;
            for(int j=0;j<lenMax;j++) {
                if(lenAA>0 && lenBB>0) {                    
                    int s=A.charAt(lenA-1-j)+B.charAt(lenB-1-j)-48-48;
                    if((s+before)<10) {
                        C[lenMax-1-j]=(char)(s+before+48);
                        before=0;
                    }else {
                        C[lenMax-1-j]=(char)(s+before-10+48);
                        before=1;
                    }
                }
            }
            System.out.println("Case "+(i+1)+":");
            System.out.print(A+" + "+B+" = ");
            if(before>0) System.out.print(before);
            System.out.println(C);
            System.out.println();
        }
    }
}

为了入门java所以都用java写，功能实现了，但OJ平台没通过，不过我也不是为了比赛，只是想练习下java，就不管那么多了，然后才发现java有BigInteger和BigDecimal类可实现任意精度的整数和浮点运算，且有专门的处理函数，啊，好方便，简直了，然后就是最后的实现。

实现

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner =new Scanner(System.in);
        int T=scanner.nextInt();
        for(int i=0;i<T;i++) {
            BigInteger A=scanner.nextBigInteger();
            BigInteger B=scanner.nextBigInteger();
            System.out.println("Case "+(i+1)+":");
            System.out.println(A+" + "+B+" = "+A.add(B));
            if(i<(T-1)) {
                System.out.println();    //格式要求
            }
        }
    }
}