1002. A+B for Polynomials (25)-优快云博客

本文探讨了如何使用队列数据结构解决多项式相加的问题，详细阐述了解题思路、关键点及实现代码。

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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 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (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

使用队列求解。
本题遇到两个易错点。
1、当两个数相加后系数为0，则不应入队。
2、两个数相加时，指数不需要相加，等于两者中任一一个即可。

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
struct node{
    int e;
    float c;
    node(){
        cin>>e>>c;
    }
};
int main(){
    queue <node> a1,a2,a;
    int k;
    cin>>k;
    while(k--)
    {
        a1.push(node());
    }
    cin>>k;
    while(k--)
    {
        a2.push(node());
    }
    while(!a1.empty()&&!a2.empty())
    {
        if(a1.front().e>a2.front().e)
        {
            node temp;
            temp.c=a1.front().c;    
            temp.e=a1.front().e;
            a.push(temp);
            a1.pop();
        }
        else if(a1.front().e<a2.front().e){
            node temp;
            temp.c=a2.front().c;    
            temp.e=a2.front().e;
            a.push(temp);
            a2.pop();
        }
        else {
            node temp;
            temp.c=a2.front().c+a1.front().c;   
            temp.e=a2.front().e;
            if(temp.c!=0 )a.push(temp);
            a1.pop();
            a2.pop();
        }
    }
    if(!a2.empty()) a1=a2;
    while(!a1.empty()){
        a.push(a1.front());
        a1.pop();
    }
    cout<<a.size();
    while(!a.empty()){
        printf(" %d %.1f",a.front().e,a.front().c);
        a.pop();
    }
    return 0;
}