[PAT甲级]1002. A+B for Polynomials (25)(求两个多项式的和)

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本文介绍了一个算法问题，即计算两个多项式的和，并提供了一段C++代码实现。输入为两个多项式的系数和指数，输出为这两个多项式相加后的结果。

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1002. A+B for Polynomials (25)

原题链接
 相似题目 1009. Product of Polynomials (25)(求多项式的积)

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：

给定两个多项式，输出它们的和
样例中多项式A有2个项： 2.4 * X^1 + 3.2 * X^0
样例中多项式B有2个项： 1.5 * X^2 + 0.5 * X^1
A+B有3个项 = 1.5 * X^2 + 2.9 * X^1 + 3.2 * X^0
X^2 表示 X的平方

代码：

#include <iostream>
#include <vector>
#include <cstdio>
using namespace std;

int main()
{
    vector<double> arr(1001);
    int A;
    cin >> A;
    for(int i=0; i<A; i++){
        int n;
        double an;
        cin >> n >> an;
        arr[n] += an;
    }
    int B;
    cin >> B;
    for(int i=0; i<B; i++){
        int n;
        double an;
        cin >> n >> an;
        arr[n] += an;
    }
    vector<int> res;
    for(int i=arr.size()-1; i>=0; i--){
        if(arr[i] != 0){
            res.push_back(i);
        }
    }
    cout << res.size();
    for(int i=0; i<res.size(); i++){
        printf(" %d %.1f", res[i], arr[res[i]]);
    }
    return 0;
}