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 N1 aN1 N2 aN2 ... NK aNK
where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤NK<⋯<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
计算多项式A+B
输入格式：
要求包含一个测试数据，每个测试数据要包含两行，每行包括一个多项式：
K N1 an1 N2 an2....NK ank
K是多项式中非零项的个数，Ni和ani是分别指数和系数。k在【1，10】，NK逐个递减【0，1000】
输出格式：
每一个测试数据，你应该在一行中输出A和B的和，和输入的格式一样，要注意在每一行的最后不要包含空格，请精确到小数点后1位。
Sample Input:
2 1 2.4 0 3.2
2 2 1.5 1 0.5
## 多项式A: 2.4*X^1 +3.2*X^0
## 多项式B: 1.5*X^2 + 0.5 *X^1
A+B = 1.5*X^2 +2.9*X^1 + 3.2
转换为输出为： 3 2 1.5 1 2.9 0 3.2
Sample Output:
3 2 1.5 1 2.9 0 3.2计算：
#include<stdio>
#include<map>
using namespace::std;
map<int,double> s;
int main()
{
int i1,t1;
double t2;
for(int z=0;z<2;z++)
{
scanf("%d",&i1);
for(int i=0;i<i1;i++)
{
scanf("%d%lf",&t1,&t2);
if(s.count(t1)==0)
s[t1]=t2;
else
s[t1]+=t2;
}
}
i1=0;
for(map<int,double>::const_iterator
m_it=s.begin();
m_it!=s.end();m_it++)
{
if(m_it->second!=0.0&&m_it->second!=-0.0)
i1++;
}
printf("%d",i1);
for(map<int,double>::reverse_iterator m_it=s.rbegin();m_it!=s.rend();m_it++)
{
if(m_it->second!=0.0&&m_it->second!=-0.0)
printf(" %d %.1lf",m_it->first,m_it->second);
}
printf("\n");
return 0;
}