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

题目详情 - 1002 A+B for Polynomials (25 分) (pintia.cn)

This time, you are supposed to find A+B where A and B are two polynomials.

Input Specification:

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 Specification:

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

 题意：计算多项式A+B的和。

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int K = 0, n = 0;
    float a = 0, N[1234] = {0};
    cin >> K;
    for(int i=1; i<=K; ++i){
        cin >> n >> a;
        N[n] += a;
    }
    cin >> K;
    for(int i=1; i<=K; ++i){
        cin >> n >> a;
        N[n] += a;
    }
    int cnt = 0;
    for(int i=0; i<=1000; ++i){
        if(N[i] != 0) ++cnt;
    }
    cout << cnt;
    for(int i=1000; i>=0; --i){
        if(N[i] != 0.0)
            printf(" %d %.1lf", i, N[i]);
    }
    return 0;
}