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 3.22 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

#include
        
         #include
         
          #include
          
           #include
           
            #include
            
             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;}
            
           
          
         
        

//写的代码有点烂。总之就是遍历呗，没想到很好的办法。就是这样下去。就是多项式对应系数相加。没什么难点。

发现了一个可以改进的地方，就是可以把第二个数组都进来的数直接相加，而不是定义两个数组。减小了空间占用。