PAT刷题 （Java语言）

1001. A+B Format (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

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

我的代码：

package _1001;

import java.text.DecimalFormat;
import java.text.NumberFormat;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        while(sc.hasNext()){
            int a=sc.nextInt();
            int b=sc.nextInt();
            int ans=a-b;
            NumberFormat format=new DecimalFormat("#,###,###");
            String str=new String();
            str=format.format(ans);
            System.out.println(str);
        }
    }
}

---

1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

package _1002;

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        int n;
        double[] a=new double[10];
        double[] b=new double[10];
        double[] ans=new double[10];
        int i;
        int max=0;
        
        //第一行
        Scanner scan=new Scanner(System.in);
        n=scan.nextInt();
        for(i=0;i<n;i++){
            int index=scan.nextInt();
            a[index]=scan.nextDouble();
        }
        
        //第二行
        scan=new Scanner(System.in);
        n=scan.nextInt();
        for(i=0;i<n;i++){
            int index=scan.nextInt();
            b[index]=scan.nextDouble();
        }
        
        for(i=0;i<10;i++){
            ans[i]=b[i]+a[i];
        }
        for(i=9;i>=0;i--){
            if(ans[i]!=0){
                max=i+1;
                break;
            }
        }
        System.out.print(max);
        for(i=max-1;i>=0;i--){//保证一位小数规格化输出
            System.out.format(" %d %.1f", i,ans[i]);
        }
        System.out.println();
    }

}