欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页

PAT甲级练习题1001、1002

程序员文章站 2022-07-15 13:50:30
...

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<A.length();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<c;i++) {
                       System.out.print(A.charAt(i));
                   }
                   if(c!=0&&len>3) System.out.print(',');
                   
                   for(int i=c;i<A.length();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:

N1​​ aN1​​​​ N2​​ aN2​​​​ ... NK​​ aNK​​​​

where K is the number of nonzero terms in the polynomial, Ni​​ and aNi​​​​ (,) 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<Integer, Double> 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