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[];

bool b[];

struct node

{

    int x;

    double y;

};

int main()

{

    int n,m;

    int max=;

    while(cin>>n)

    {

        memset(a,,sizeof(a));

        memset(b,,sizeof(b));

        int s=;

        max=;

        for(int i=;i<=n;i++)

        {

            int x;

            double y;

            cin>>x>>y;

            a[x]+=y;

            if(x>max) max=x;

            if(!b[x])

            {

                b[x]=;

            }

        }

        cin>>m;

        for(int i=;i<=m;i++)

        {

            int x;

            double y;

            cin>>x>>y;

            a[x]+=y;

            if(x>max) max=x;

            if(!b[x])

            {

                b[x]=;

            }

        }

        queue<node>q;

        while(!q.empty ()) q.pop();

        for(int i=max;i>=;i--)

        {

            if(a[i]!=)

            {

                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 ;

}

巴特西

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

1002 A+B for Polynomials (25)（25 分）

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