PAT甲级练习题1001、1002

PAT甲级练习题1001、1002

1001 A+B Format (20 分)

 

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

-1000000 9

Sample Output:

-999,991


解题时注意取余时i的位置
满分代码：

import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNextInt()) {
            int a = in.nextInt();
            int b = in.nextInt();
            int res = a+b;
            String A = String.valueOf(res);
               if(res<0) {
                   int len = A.length();
                   int c = (len-1)%3;
                   System.out.print('-');
                   for(int i=1;i<=c;i++) {
                       System.out.print(A.charAt(i));
                   }
                   if(c!=0&&len>3) System.out.print(',');
                   
                   for(int i=c+1;i) {
                       if((i-c-1)!=0&&(i-c-1)%3==0&&i!=A.length()-1) System.out.print(',');
                       System.out.print(A.charAt(i));
                   }
                   System.out.println();
               }else {
                   int len = A.length();
                   int c = len%3;
                   for(int i=0;i) {
                       System.out.print(A.charAt(i));
                   }
                   if(c!=0&&len>3) System.out.print(',');
                   
                   for(int i=c;i) {
                       if((i-c)!=0&&(i-c)%3==0&&i!=A.length()-1) System.out.print(',');
                       System.out.print(A.charAt(i));
                   }
                   System.out.println();
               }
        }
        in.close();
    }
}

1002 A+B for Polynomials (25 分)

 

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

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:

K N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (,) are the exponents and coefficients, respectively. It is given that 1，0.

Output Specification:

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

注意为0项需要溢出map表
满分代码：

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNextInt()) {
            int keyMax = 0;
            int up = in.nextInt();
            Map hm = new HashMap<>();
            for (int i = 0; i < up; i++) {
                int x = in.nextInt();
                if (x > keyMax)
                    keyMax = x;
                hm.put(x, in.nextDouble());
            }
            int down = in.nextInt();
            for (int i = 0; i < down; i++) {
                int x = in.nextInt();
                if (x > keyMax)
                    keyMax = x;

                if (!hm.containsKey(x)) {
                    hm.put(x, in.nextDouble());
                } else {
                    double y = hm.get(x);
                    y += in.nextDouble();
                    if(y!=0)
                    hm.put(x, y);
                    else hm.remove(x);
                }
            }
            if(hm.size()==0) System.out.print("0");
            else
                System.out.print(hm.size() + " ");
            int sign = 0;
            for (int i = keyMax; i >= 0; i--) {
                if (hm.containsKey(i)) {
                    sign++;
                        
                    System.out.print(i + " ");
                    if (sign != hm.size())
                        System.out.print(String.format("%.1f", hm.get(i)) + " ");
                    else
                        System.out.print(String.format("%.1f", hm.get(i)));
                }
            }
        }
    }
}

 

posted @ 2019-05-20 16:53 godoforange 阅读( ...) 评论( ...) 编辑 收藏