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<) {
int len = A.length();
int c = (len-)%;
System.out.print('-');
for(int i=;i<=c;i++) {
System.out.print(A.charAt(i));
}
if(c!=&&len>) System.out.print(',');
for(int i=c+;i<A.length();i++) {
if((i-c-)!=&&(i-c-)%==&&i!=A.length()-) System.out.print(',');
System.out.print(A.charAt(i));
}
System.out.println();
}else {
int len = A.length();
int c = len%;
for(int i=;i<c;i++) {
System.out.print(A.charAt(i));
}
if(c!=&&len>) System.out.print(',');
for(int i=c;i<A.length();i++) {
if((i-c)!=&&(i-c)%==&&i!=A.length()-) 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 = ;
int up = in.nextInt();
Map<Integer, Double> hm = new HashMap<>();
for (int i = ; i < up; i++) {
int x = in.nextInt();
if (x > keyMax)
keyMax = x;
hm.put(x, in.nextDouble());
}
int down = in.nextInt();
for (int i = ; 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!=)
hm.put(x, y);
else hm.remove(x);
}
}
if(hm.size()==) System.out.print("");
else
System.out.print(hm.size() + " ");
int sign = ;
for (int i = keyMax; i >= ; 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)));
}
}
}
}
}
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