PAT_1002.A+B for Polynomials

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本文介绍了一种解决多项式相加问题的算法实现方法，输入包含两个多项式的非零项系数及其指数，输出为这两个多项式的和。通过使用C++语言进行编程，实现了多项式的读取、加法运算及结果输出。

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//1002. A+B for Polynomials (25)
//
//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

#include "stdafx.h"
#include "stdio.h"
#include "iostream"
#include "string.h"
#include "stdlib.h"
#include "string"
#include "algorithm"
using namespace std;

float a[1010],b[1010],c[1010];
int main()
{
    int k1,k2;
    while(cin>>k1){
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i =0;i<k1;i++){
            int j;
            cin>>j;
            cin>>a[j];
        }
        cin>>k2;
        for(int i = 0;i<k2;i++){
            int k;
            cin>>k;
            cin>>b[k];
        }
        int count = 0;
        for(int i = 1002;i>=0;i--){
            c[i] = a[i]+b[i];
            if(c[i])
                count++;
        }
        cout<<count;
        for(int i = 1002;i>=0;i--){
            if(c[i])
                printf(" %d %.1lf",i,c[i]);
                //cout<<" "<<i<<" "<<c[i];
        }
        cout<<endl;
    }
    return 0;
}