PAT 甲级 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 N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

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

代码： 

#include<stdio.h>
#include<string>
#include<queue>
using namespace std;
struct node{
	double coef;//系数
	int exp;//指数
};


int main(){
	int num;
	queue<node> p1,p2,sum;
	if (scanf("%d", &num) != -1){
		for (int i = 0; i < num; i++){
			node temp;
			if ((scanf("%d", &temp.exp) != -1)){
				if (scanf("%lf", &temp.coef) != -1){
					p1.push(temp);
					//printf("aasa%d,,%d,,,,", temp.exp, temp.coef);
				}
			}
		}
		//printf("asadas\n");
	}
	if (scanf("%d", &num) != -1){
		//node p1[num];
		for (int i = 0; i < num; i++){
			node temp;
			if (scanf("%d", &temp.exp) != -1){
				if (scanf("%lf", &temp.coef) != -1){
					p2.push(temp);
					//printf("aasa%d,,%d,,,,", temp.exp, temp.coef);
				}
			}
		}
	}
	while (p1.empty() == false && p2.empty()==false){
		node temp1 = p1.front();
		node temp2 = p2.front();
		if (temp1.exp > temp2.exp){
			p1.pop();
			sum.push(temp1);
		}
		else if (temp1.exp < temp2.exp){
			p2.pop();
			sum.push(temp2);
		}
		else{
			p1.pop();
			p2.pop();
			node temp;
			temp.exp = temp1.exp;
			temp.coef = temp1.coef + temp2.coef;
			if (temp.coef != 0){
				sum.push(temp);
			}
		}
	}
	//printf("111");
	while (p1.empty() == false){
		node temp = p1.front();
		p1.pop();
		sum.push(temp);
	}
	//printf("222");
	while (p2.empty() == false){
		node temp = p2.front();
		p2.pop();
		sum.push(temp);
	}
	printf("%d", (int)sum.size());
	while (sum.empty() == false){
		node temp = sum.front();
		sum.pop();
		printf(" %d %.1f", temp.exp, temp.coef);
	}
	


	return 0;
}

思路：

输入的多项式是按照指数大小兄大到小输入的，是有序的，因此可以参考归并排序合并两个有序数组的思路将对多项式求和。

同样，若输入的多项式无序，可以先排序后按同样的方法处理。

易错点：

两个多项式求和后的结果若为0，不要将其加入结果多项式中。