PAT 甲级 1002 A+B for Polynomials（25）

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 N​_1​​ a_​N​1​​​​ N​₂​​ a​_N​2​​​​ ... N​_K​​ a​_N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1 ≤ K ≤ 10，0 ≤ N​_K ​​ < ⋯ < N_​2​​ < N​_1​​ ≤ 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

Experiential Summing-up

两种办法：

①map（较麻烦）：使用三个 map ，键值对为<指数, 系数>，最后利用 vector 进行排序输出。

②array：使用一个数组，将指数相同的系数相加，统计非零系数的个数，最后倒序输出系数不为 0 的指数和系数。

Accepted Code

#include<bits/stdc++.h>
using namespace std;

int k, N;
double a;
vector<int> x;
map<int, double> map1, map2, map3;

int main()
{
    cin >> k;
    for(int i = 1; i <= k; i ++)
    {
        cin >> N >> a;
        map1[N] = a;
    }
    cin >> k;
    for(int i = 1; i <= k; i ++)
    {
        cin >> N >> a;
        map2[N] = a;
    }
    for(int i = 0; i <= 1000; i ++)
    {
        if(map1[i] + map2[i] != 0)
        {
            x.push_back(i);
            map3[i] = map1[i] + map2[i];
        }
    }

    sort(x.begin(), x.end());
    reverse(x.begin(), x.end());

    cout << x.size();
    for(int i = 0; i < x.size(); i ++)
    {
        cout << " " << x[i] << " ";
        cout << fixed <<setprecision(1) <<map3[x[i]];
    }
    return 0;
}

 

#include<bits/stdc++.h>
using namespace std;

int n, m, t;
int cnt = 0;
float num;
float c[1001];

int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; i ++)
    {
        scanf("%d%f", &t, &num);
        c[t] += num;
    }
    scanf("%d", &m);
    for(int i = 0; i < m; i ++)
    {
        scanf("%d%f", &t, &num);
        c[t] += num;
    }

    for(int i = 0; i < 1001; i ++)
        if(c[i] != 0) cnt ++;
    printf("%d", cnt);

    for(int i = 1000; i >= 0; i --)
    {
        if(c[i] != 0.0)
            printf(" %d %.1f", i, c[i]);
    }
    return 0;
}