1002. A+B for Polynomials (25)

最新推荐文章于 2021-09-07 12:21:13 发布

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#Polynomials #A+B

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本文介绍了一种多项式加法的实现方法，通过解析输入文件中的多项式表达式，进行相加运算，并按指定格式输出结果。输入包含两个多项式，每个多项式由若干项组成，包括指数和系数。

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

多项式相加,输入一共2行,每行第一个表示项数,sample表示 第一个多项式有2项,为 2.4x+3.2,第二个多项式为两项,为 1.5x^2+0.5x,两个多项式相加以后为3项,为1.5x^2+2.9x+3.2,输入的一个多项式最多10项,但多项式最高幂为1000,还有系数保留一位小数;

//
//  main.cpp
//  PAT
//
//  Created by hqr on 14-2-7.
//  Copyright (c) 2014年 hqr. All rights reserved.
//

#include <iostream>
using namespace std;

int nA,nB,nC=0;
double input;
double fAns[1001];

int main(int argc, const char * argv[])
{

    // insert code here...
    cin >> nA;
    
    for(int i=0;i<nA;i++)
    {
        cin>>nB>>input;
        fAns[nB]+=input;
    }
    
    cin >> nA;
    
    for(int i=0;i<nA;i++)
    {
        cin>>nB>>input;
        fAns[nB]+=input;
    }
    
    for(int i=0;i<=1000;i++)
    {
        if (fAns[i]!=0) {
            nC++;
        }
        
    }
    
    cout << nC;
    
    for(int i=1000;i>=0;i--)
    {
        if (fAns[i]!=0) {
            printf(" %d %.1lf",i,fAns[i]);
        }
        
    }
    cout<<endl;
    
    
    return 0;
}