1002. A+B for Polynomials (25)

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

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		double s[]=new double [1001];
		for(int j=1000;j>=0;j--){
			s[j]=0;
			}
		Scanner in=new Scanner(System.in);
		int i=1,k;
		while(i<3){
			k=Integer.parseInt(in.next());
			for(int j=0;j<k;j++){
				int n=Integer.parseInt(in.next());
				s[n]=s[n]+Double.parseDouble(in.next());
			}
			i++;
		}
		int count=0;
		for(double t:s){
			if(t!=0) count++;
		}
		System.out.print(count);
		k=0;
		for(int j=1000;j>=0;j--){
			if(s[j]-0>0&&k<count){
				System.out.printf(" %d %.1f",j,s[j]);
				k++;
			}
		}
		System.out.println();
	}

}

　　提交结果部分正确

修改后

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		double s[]=new double [1001];
		for(int j=1000;j>=0;j--){
			s[j]=0;
			}
		Scanner in=new Scanner(System.in);
		int i=1,k;
		while(i<3){
			k=Integer.parseInt(in.next());
			for(int j=0;j<k;j++){
				int n=Integer.parseInt(in.next());
				s[n]=s[n]+Double.parseDouble(in.next());
			}
			i++;
		}
		int count=0;
		for(double t:s){
			if(t!=0) count++;
		}
		System.out.print(count);
		k=0;
		for(int j=1000;j>=0;j--){
			if(Math.abs(s[j]-0)>0&&k<count){
				System.out.printf(" %d %.1f",j,s[j]);
				k++;
			}
		}
		System.out.println();
	}

}

　　提交结果

或者使用DOuble.compare(double,double)比较

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		double s[]=new double [1001];
		for(int j=1000;j>=0;j--){
			s[j]=0;
			}
		Scanner in=new Scanner(System.in);
		int i=1,k;
		while(i<3){
			k=Integer.parseInt(in.next());
			for(int j=0;j<k;j++){
				int n=Integer.parseInt(in.next());
				s[n]=s[n]+Double.parseDouble(in.next());
			}
			i++;
		}
		int count=0;
		for(double t:s){
			if(t!=0) count++;
		}
		System.out.print(count);
		k=0;
		for(int j=1000;j>=0;j--){
			if(Double.compare(s[j],0.0)!=0&&k<count){
				System.out.printf(" %d %.1f",j,s[j]);
				k++;
			}
		}
		System.out.println();
	}

}

　其实s[j]!=0.0也可以，所以大概是要注意doubel计算中的精度问题　

参考

http://blog.csdn.net/u014646950/article/details/46932525