1001. A+B Format (20)

時間限制 400 ms 內(nèi)存限制 65536 kB 代碼長度限制 16000 B 判題程序 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.

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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

我的代碼： 1 package _1001; 2 3 import java.text.DecimalFormat; 4 import java.text.NumberFormat; 5 import java.util.Scanner; 6 7 public class Main { 8 9 public static void main(String[] args) { 10 Scanner sc=new Scanner(System.in); 11 while(sc.hasNext()){ 12 int a=sc.nextInt(); 13 int b=sc.nextInt(); 14 int ans=a-b; 15 NumberFormat format=new DecimalFormat("#,###,###"); 16 String str=new String(); 17 str=format.format(ans); 18 System.out.println(str); 19 } 20 } 21 }

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1002. A+B for Polynomials (25)

時間限制 400 ms 內(nèi)存限制 65536 kB 代碼長度限制 16000 B 判題程序 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.

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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 1 package _1002; 2 3 import java.util.Scanner; 4 5 public class Main { 6 7 public static void main(String[] args) { 8 9 int n; 10 double[] a=new double[10]; 11 double[] b=new double[10]; 12 double[] ans=new double[10]; 13 int i; 14 int max=0; 15 16 //第一行 17 Scanner scan=new Scanner(System.in); 18 n=scan.nextInt(); 19 for(i=0;i<n;i++){ 20 int index=scan.nextInt(); 21 a[index]=scan.nextDouble(); 22 } 23 24 //第二行 25 scan=new Scanner(System.in); 26 n=scan.nextInt(); 27 for(i=0;i<n;i++){ 28 int index=scan.nextInt(); 29 b[index]=scan.nextDouble(); 30 } 31 32 for(i=0;i<10;i++){ 33 ans[i]=b[i]+a[i]; 34 } 35 for(i=9;i>=0;i--){ 36 if(ans[i]!=0){ 37 max=i+1; 38 break; 39 } 40 } 41 System.out.print(max); 42 for(i=max-1;i>=0;i--){//保證一位小數(shù)規(guī)格化輸出 43 System.out.format(" %d %.1f", i,ans[i]); 44 } 45 System.out.println(); 46 } 47 48 }

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轉(zhuǎn)載于:https://www.cnblogs.com/TQCAI/p/7718778.html

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