java_patest甲级真题1002. A+B for Polynomials

This time, you are supposed to find A+B where A and B are two polynomials.

Input

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, where
K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < ... < N2 < N1 <=1000.

Output

For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.
Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner scanner = new Scanner(System.in);
String str1 = scanner.nextLine();
String str2 = scanner.nextLine();
String[] temp1 = str1.split(" ");
String[] temp2 = str2.split(" ");
float[] fA = new float[temp1.length] ;
float[] fB = new float[temp2.length];
for(int i = 0 ; i<temp1.length; i++)
{
fA[i] = Float.parseFloat(temp1[i]);
}
for(int i = 0 ; i<temp2.length; i++)
{
fB[i] = Float.parseFloat(temp2[i]);
}
boolean flag = true;
int i = 1 , j = 1;
StringBuilder sb = new StringBuilder("");
String str = "";
float temp = 0;
int count = 0;
while(flag)
{
if((i == fA.length)&&(j == fB.length))
break;
if((i == fA.length)&&(j < fB.length))
{
sb.append((" "+(int)fB[j]+" "+fB[j+1]));
j = j +2;
count++;
continue ;
}
if((i < fA.length)&&(j == fB.length))
{
sb.append(" "+(int)fA[i]+" "+fA[i+1]);
i = i +2;
count++;
continue ;
}
if((i< fA.length)&&(j<fB.length))
if((int)fA[i] == (int)fB[j])
{
str = " ";
temp = fA[i+1]+fB[j+1];
if(temp == 0.0)
{
i = i+2;
j = j+2;
continue;
}
str = str +(int)fA[i]+" "+temp;
sb.append(str);
i = i +2;
j = j +2;
}else if(fA[i]>fB[j])
{
sb.append((" "+(int)fA[i]+" "+fA[i+1]));
i = i +2;
}else
{
sb.append((" "+(int)fB[j]+" "+fB[j+1]));
j = j +2 ;
}
count++;
}
sb.insert(0,count);
String string = new String(sb);
String[] s = string.split(" ");
System.out.print(Integer.parseInt(s[0]));
for(int k =1 ; k< s.length ;k +=2)
{
System.out.print(" "+s[k] +" ");
System.out.printf("%.1f",Float.parseFloat(s[k+1]));
}
}
}


本题分析： 这道题要求是 A B两个多项式想加，一道很简单的题，但是有个坑 神坑···如果两个数相加 系数刚好抵消掉，那就呵呵了。 所以被这个坑了起码四五次。

当知道了原因后 就好办。 题目分析： 首先 这道题 A B 两个 都是 有顺序的 N值是递减的，所以 每次比较都可以先比较下，如果A的大 那么就把A 加进去，然后A的指针往后移动，如果B 那就加B，如果相同，先把数值相加·然后加进去，当然 如果相同的情况还要判断 是否为0 ；

其次 ： 

重要： 这道题 可以还有另外两种，第一种： 看这题··首先想到 如果用容器做的话 ，第一 用MAP 可以实现，第二 用list 的话 那么只要构建对象就能实现。