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:
K 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))); } } } } }