PAT甲级1002多项式相加

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

题目大意

计算多项式A+B的和

分析

设立c数组，长度为指数的最大值，c[i] = j表示指数i的系数为j，接收a和b输入的同时将对应指数的系数加入到c中，累计c中所有非零系数的个数，然后从后往前输出所有系数不为0的指数和系数

Code

//
//  PAT 甲 1002A+B多项式.cpp
//  Fushi
//
//  Created by H C on 2021/2/5.
//

#include<stdio.h>
using namespace std;

double a[1010]={0};
double b[1010]={0};
double c[1010]={0};

int main()
{
    int K1,K2;
    scanf("%d",&K1);
    while (K1--) {
        int N;
        double A;
        scanf("%d %lf",&N,&A);
        a[N] = A;
    }
    scanf("%d",&K2);
    while (K2--) {
        int N;
        double B;
        scanf("%d %lf",&N,&B);
        b[N] = B;
    }
    for (int i=0; i<1010; i++)
    {
        if (a[i]!=0||b[i]!=0) {
            c[i] = a[i]+b[i];
        }
    }
    int count=0;
    for (int i=1010; i>=0; i--) {
        if(c[i]!=0) count++;
    }
    printf("%d",count);
    for (int i=1010; i>=0; i--) {
        if (c[i]!=0) {
            printf(" %d %.1lf",i,c[i]);
        }
    }
}