PAT甲级 1002 A+B for Polynomials

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:

$K$K $N_1$N1​ $a_{N_1}$aN1​​ $N_2$N2​ $a_{N_2}$aN2​​ $...$... $N_K$NK​ $a_{N_K}$aNK​​
​​
where K is the number of nonzero terms in the polynomial, $N_​i$N​​i and $a_{​N​_i}$a​N​i​​
​​
​​ $(i=1,2,⋯,K)$(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.$1≤K≤10，0≤N​K​<⋯<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

多项式的A+B，符合PAT的套路，坑多，注意相加也会出现0的情况，直接暴力算即可，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e3+5;
double p[maxn]={0.0};
int k, n, ans = 0;
double a;
int main() {
    scanf("%d", &k);
    for(int i = 0; i < k; i++) {
        scanf("%d %lf", &n, &a);
        p[n] += a;
    }
    scanf("%d", &k);
    for(int i = 0; i < k; i++) {
        scanf("%d %lf", &n, &a);
        p[n] += a;
    }
    for(int i = 0; i < maxn; i++) {
        if(p[i] != 0) {
            ans++;
        }
    }
    printf("%d", ans);
    for(int i = maxn; i >= 0; i--) {
        if(p[i] != 0) {
            printf(" %d %.1f", i, p[i]);
        }
    }
    return 0;
}

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