1002. A+B for Polynomials (25)

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本文介绍了一种解决多项式加法问题的算法实现方法，输入两个多项式A和B，输出它们相加的结果。该算法通过读取每个多项式的项数及各项系数和指数，进行合并计算并输出结果。

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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<list>
using namespace std;
struct node{
int ex;
double coe;
}L1[20],L2[20],L3[20];
int main()
{
//freopen("in.txt","r",stdin);
int len1,len2;
int i=0;
cin>>len1;

for(i=0;i<len1;i++)
{
cin>>L1[i].ex;
cin>>L1[i].coe;

}
cin>>len2;

int j;
for(j=0;j<len2;j++)
{
cin>>L2[j].ex;
cin>>L2[j].coe;

}
int k=0;

for(i=0,j=0;i<len1&&j<len2;)
{

if(L1[i].ex>L2[j].ex)
{
L3[k].ex=L1[i].ex;
L3[k].coe=L1[i].coe;
k++;
i++;

}
else if(L1[i].ex<L2[j].ex)
{
L3[k].ex=L2[j].ex;
L3[k].coe=L2[j].coe;
k++;
j++;

}
else
{
if((L2[j].coe+L1[i].coe)!=0)
{
L3[k].ex=L2[j].ex;
L3[k].coe=L2[j].coe+L1[i].coe;


k++;
}
i++;
j++;

}



}

while(i<len1)
{

L3[k].ex=L1[i].ex;
L3[k].coe=L1[i].coe;
k++;
i++;


}
while(j<len2)
{

L3[k].ex=L2[j].ex;
L3[k].coe=L2[j].coe;
k++;
j++;
}

if(k>0)
{
cout<<k<<" ";
for(i=0;i<k-1;i++)
{
printf("%d %.1lf ",L3[i].ex,L3[i].coe);

}
printf("%d %.1lf",L3[k-1].ex,L3[k-1].coe);
}
else
cout<<k;
cout<<endl;

return 0;
}