PAT甲级 1002 A+B for Polynomials (25)（25 分）

1002 A+B for Polynomials (25)（25 分）

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

1002.多项式A与B的和

这次，假设A和B是两个多项式，求A与B的和多项式。

 

输入

每个输入文件包含一个测试实例。每个实例有两行，每行包含一个多项式的信息: K N1 a_N1 N2 a_N2 ... NK a_NK，其中K为多项式中非0项的个数，Ni 和 a_Ni (i=1, 2, ..., K) 分别为指数和系数。数的范围是1 <= K <= 10，0<= NK < ... < N2 < N1 <=1000。

输出

对于每个测试实例，你需要在一行内输出A与B的和，格式与输入时相同。注意每行的结尾不能有多余的空格。小数精确到一位。

多项式求和
可能会有负数

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
double a[1005];
bool b[1005];
struct node
{
    int x;
    double y;
};
int main()
{
    int n,m;
    int max=0;
    while(cin>>n)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        int s=0;
        max=0;
        for(int i=1;i<=n;i++)
        {
            int x;
            double y;
            cin>>x>>y;
            a[x]+=y;
            if(x>max) max=x;
            if(!b[x])
            {
                b[x]=1;
            }
        }
        cin>>m;
        for(int i=1;i<=m;i++)
        {
            int x;
            double y;
            cin>>x>>y;
            a[x]+=y;
            if(x>max) max=x;
            if(!b[x])
            {
                b[x]=1;
            }
        }
        queue<node>q;
        while(!q.empty ()) q.pop();
        for(int i=max;i>=0;i--)
        {
            if(a[i]!=0)
            {
                node p;
                p.x=i;
                p.y=a[i];
                q.push (p);
                s++;
            }
        }
        cout<<s;
        while(!q.empty ())
        {
            node p=q.front();
            q.pop();
            printf(" %d %.1lf",p.x,p.y);
        }
        cout<<endl;
        
    }
    return 0;

}