1002 A+B for Polynomials (25)（25 分）

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 a~N1~ N2 a~N2~ ... NK a~NK~, where K is the number of nonzero terms in the polynomial, Ni and a~Ni~ (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

#include<iostream>
#include
 
<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
double nk1[1001]={0},nk2[1001]={0};
int main()
{
    //freopen("1.txt","r",stdin);
    int n;
    scanf("%d",&n);
    int a;
    double b;
    while(n--){
        scanf("%d%lf",&a,&b);
        nk1[a]
 
=b;
    }
    scanf("%d",&n);
    while(n--){
        scanf("%d%lf",&a,&b);
        nk2[a]=b;
    }

    for(int i=0;i<1001;i++){
        nk1[i]=nk1[i]+nk2[i];
    }
    int ct=0;
    for(int i=0;i<1001;i++){
        if(nk1[i]!=0){
            ct++;
        }
    }
    printf("%d",ct);
    if(ct!=0)printf(" ");
    for(int i=1000;i>=0;i--){
        if(nk1[i]!=0)
        {printf("%d %.1f",i,nk1[i]);
        ct--;
        if(ct!=0)printf(" ");
        if(ct==0)break;
        }
    }

    return 0;
}

//寫的代碼有點爛。總之就是遍歷唄，沒想到很好的辦法。就是這樣下去。就是多項式對應系數相加。沒什麽難點。

發現了一個可以改進的地方，就是可以把第二個數組都進來的數直接相加，而不是定義兩個數組。減小了空間占用。

PAT A+B for Polynomials[簡單]