大数据之大数相加（正数即可）

描述
acmj最近发现在使用计算器计算高精度的大数加法时很不方便，于是他想着能不能写个程序把这个问题给解决了。
输入
每组数据包含两个正数A,B（可能为小数且位数不大于400）
输出
每组输出数据占一行，输出A+B的结果，结果需要是最简的形式。
样例输入

1.9 0.1

样例输出

2

代码如下：

import java.util.Scanner;

/** * Created by lcc on 2017/6/5. */
public class ABProblemIV {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        String str1 = input.next();
        String str2 = input.next();
        String[] str1a = str1.split("."); //这样就变成2个大数相加的问题了
        String[] str2b = str2.split(".");

        int[] zhengshu = dashuxiangjia(str1a[0],str2b[0]);
        int[] xiaoshu = xiaoshuxiangjia(str1a[1],str2b[1]);
        int changdu1 = str1a[1].length() - str2b[1].length() > 0 ? str1a[1].length() : str2b[1].length();
        int changdu0 = str1a[0].length() - str2b[0].length() > 0 ? str1a[0].length() : str2b[0].length();
        if (xiaoshu[0] != 0) {
            zhengshu[0] = zhengshu[0] + xiaoshu[0];   //小数相加最多增加一位 如果第一位不位0，则说明小数相加>1， 将该值与整数第一位
        }
        while (zhengshu[changdu0] == 0) {
            changdu0--;
        }
        for (int i1 = changdu0; i1 >= 0; i1--) {
            System.out.print(zhengshu[i1]);
        }
        System.out.print(".");
        for (int i = 1; i < xiaoshu.length; i++) {
            System.out.print(xiaoshu[i]);
        }
    }

    /** * 整数相加 */
    public static int[] dashuxiangjia(String a,String b) {
        int changdu = a.length() - b.length() > 0 ? a.length() : b.length();
        int[] aint = new int[changdu];
        int[] bint = new int[changdu];
        for (int i1 = 0; i1 < a.length(); i1++) {
            aint[a.length() - 1 - i1] = Integer.parseInt(a.substring(i1,i1 + 1));
        }
        for (int i1 = 0; i1 < b.length(); i1++) {
            bint[b.length() - 1 - i1] = Integer.parseInt(b.substring(i1,i1 + 1));
        }
        int[] sum = new int[changdu + 1];
        for (int i3 = 0; i3 < changdu; i3++) {
            sum[i3] = aint[i3] + bint[i3];
        }
        for (int i4 = 0; i4 < changdu; i4++) {
            sum[i4 + 1] = sum[i4] / 10 + sum[i4 + 1];
            sum[i4] = sum[i4] % 10;
        }
        int sumchangdu = changdu;
/* while (sum[sumchangdu] == 0) { sumchangdu--; }*/
        return sum;
/* System.out.print(a + " + " + b + " = "); for (int i5 = sumchangdu; i5 >= 0; i5--) { System.out.print(sum[i5]); }*/
    }

    /** * 小數相加 * * @param a * @param b * @return */
    public static int[] xiaoshuxiangjia(String a,String b) {
        int changdu = a.length() - b.length() > 0 ? a.length() : b.length();
        int[] aint = new int[changdu + 1];
        int[] bint = new int[changdu + 1];
        for (int i1 = 0; i1 < a.length(); i1++) {
            aint[i1 + 1] = Integer.parseInt(a.substring(i1,i1 + 1));
        }
        for (int i1 = 0; i1 < b.length(); i1++) {
            bint[i1 + 1] = Integer.parseInt(b.substring(i1,i1 + 1));
        }
        int[] sum = new int[changdu + 1];
        for (int i3 = 0; i3 <= changdu; i3++) {
            sum[i3] = aint[i3] + bint[i3];
        }
        for (int i4 = changdu; i4 >= 1; i4--) {
            sum[i4 - 1] = sum[i4] / 10 + sum[i4 - 1];
            sum[i4] = sum[i4] % 10;
        }
/* while (sum[sumchangdu] == 0) { sumchangdu--; }*/
        return sum;
/* System.out.print(a + " + " + b + " = "); for (int i5 = sumchangdu; i5 >= 0; i5--) { System.out.print(sum[i5]); }*/
    }


}